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Here is an example with Cohen forcing. Let $\mathbb{Q}$ be the forcing that adds two Cohen reals, viewed as binary sequences, along with their bitwise sum mod 2. So conditions are binary sequences $(p,q,r)$ all of the same length, such that $r=p\oplus q$, ordered by extension.

This forcing will add a Cohen real on each coordinate (since the sum of two Cohen reals is also generic for Cohen forcing). And so we will get $M[G_0,G_1,G_2]$ above $M[G_0]$, $M[G_1]$, and $M[G_2]$, and any two of them will be mutually generic, so we will have $M[G_i]\cap M[G_j]=M$ for $i\neq j$. But once you have two of them, you can define the third, and so you get the whole thing.

Of course, this picture isn't the complete lattice of grounds, since every Cohen real can be split in two, and so if you had meant to ask about that one would need a more precise question.

Meanwhile, you ask about all possible $\mathbb{P}_i$, but the answer in this generality is false.

Here is one quick way to make an example. Take a Suslin tree $T$ that is $3$-fold Suslin off the generic branch. This is a strong form of rigidity, which means that after forcing with the tree, what remains above any node not on the generic branch is still Suslin. And again.

If we have any three generics for this forcing, then they will be automatically mutually generic, since they meet every level of the tree and hence all maximal antichains. And so we can combine any two to form the model $M[G_0,G_1]$, which will not have $G_2$, and so it won't conform with your lattice.

Here is an example with Cohen forcing. Let $\mathbb{Q}$ be the forcing that adds two Cohen reals, viewed as binary sequences, along with their bitwise sum mod 2. So conditions are binary sequences $(p,q,r)$ all of the same length, such that $r=p\oplus q$, ordered by extension.

We can present this in a more symmetric manner as: conditions are triples of finite binary sequences $(p,q,r)$ with all the same length, such that $p\oplus q\oplus r=0$. Each of the three is the bitwise sum of the other two.

This forcing will add a Cohen real on each coordinate. And so we will get $M[G_0,G_1,G_2]$ above $M[G_0]$, $M[G_1]$, and $M[G_2]$, and any two of them will be mutually generic, so we will have $M[G_i]\cap M[G_j]=M$ for $i\neq j$. But once you have two of them, you can define the third, and so you get the whole thing.

Of course, this picture isn't the complete lattice of grounds, since every Cohen real can be split in two, and so if you had meant to ask about that one would need a more precise question.

Meanwhile, you ask about all possible $\mathbb{P}_i$, but the answer in this generality is false.

Here is one quick way to make an example. Take a Suslin tree $T$ that is $3$-fold Suslin off the generic branch. This is a strong form of rigidity, which means that after forcing with the tree, what remains above any node not on the generic branch is still Suslin. And again.

If we have any three generics for this forcing, then they will be automatically mutually generic, since they meet every level of the tree and hence all maximal antichains. And so we can combine any two to form the model $M[G_0,G_1]$, which will not have $G_2$, and so it won't conform with your lattice.

Here is an example with Cohen forcing. Let $\mathbb{Q}$ be the forcing that adds two Cohen reals, viewed as binary sequences, along with their bitwise sum mod 2. So conditions are binary sequences $(p,q,r)$ all of the same length, such that $r=p\oplus q$, ordered by extension.

This forcing will add a Cohen real on each coordinate (since the sum of two Cohen reals is also generic for Cohen forcing). And so we will get $M[G_0,G_1,G_2]$ above $M[G_0]$, $M[G_1]$, and $M[G_2]$, and any two of them will be mutually generic, so we will have $M[G_i]\cap M[G_j]=M$ for $i\neq j$. But once you have two of them, you can define the third, and so you get the whole thing.

Of course, this picture isn't the complete lattice of grounds, since every Cohen real can be split in two, and so if you had meant to ask about that one would need a more precise question.

Meanwhile, you ask about all possible $\mathbb{P}_i$, but the answer in this generality is false.

Here is one quick way to make an example. Take a Suslin tree $T$ that is $3$-fold Suslin off the generic branch. This is a strong form of rigidity, which means that after forcing with the tree, what remains above any node not on the generic branch is still Suslin. And again.

If we have any three generics for this forcing, then they will be automatically mutually generic. And so we can combine any two to form the model $M[G_0,G_1]$, which will not have $G_2$, and so it won't conform with your lattic.

Here is an example with Cohen forcing. Let $\mathbb{Q}$ be the forcing that adds two Cohen reals, viewed as binary sequences, along with their bitwise sum mod 2. So conditions are binary sequences $(p,q,r)$ all of the same length, such that $r=p\oplus q$, ordered by extension.

This forcing will add a Cohen real on each coordinate (since the sum of two Cohen reals is also generic for Cohen forcing). And so we will get $M[G_0,G_1,G_2]$ above $M[G_0]$, $M[G_1]$, and $M[G_2]$, and any two of them will be mutually generic, so we will have $M[G_i]\cap M[G_j]=M$ for $i\neq j$. But once you have two of them, you can define the third, and so you get the whole thing.

Of course, this picture isn't the complete lattice of grounds, since every Cohen real can be split in two, and so if you had meant to ask about that one would need a more precise question.

Meanwhile, you ask about all possible $\mathbb{P}_i$, but the answer in this generality is false.

Here is one quick way to make an example. Take a Suslin tree $T$ that is $3$-fold Suslin off the generic branch. This is a strong form of rigidity, which means that after forcing with the tree, what remains above any node not on the generic branch is still Suslin. And again.

If we have any three generics for this forcing, then they will be automatically mutually generic, since they meet every level of the tree and hence all maximal antichains. And so we can combine any two to form the model $M[G_0,G_1]$, which will not have $G_2$, and so it won't conform with your lattice.

Here is an example. Let $\mathbb{Q}$ be the forcing that adds two Cohen reals, viewed as binary sequences, along with their bitwise sum mod 2. So conditions are binary sequences $(p,q,r)$ all of the same length, such that $r=p\oplus q$, ordered by extension.

This forcing will add a Cohen real on each coordinate (since the sum of two Cohen reals is also generic for Cohen forcing). And so we will get $M[G_0,G_1,G_2]$ above $M[G_0]$, $M[G_1]$, and $M[G_2]$, and any two of them will be mutually generic, so we will have $M[G_i]\cap M[G_j]=M$ for $i\neq j$. But once you have two of them, you can define the third, and so you get the whole thing.

Of course, this picture isn't the complete lattice of grounds, since every Cohen real can be split in two, and so if you had meant to ask about that one would need a more precise question.

Here is an example with Cohen forcing. Let $\mathbb{Q}$ be the forcing that adds two Cohen reals, viewed as binary sequences, along with their bitwise sum mod 2. So conditions are binary sequences $(p,q,r)$ all of the same length, such that $r=p\oplus q$, ordered by extension.

This forcing will add a Cohen real on each coordinate (since the sum of two Cohen reals is also generic for Cohen forcing). And so we will get $M[G_0,G_1,G_2]$ above $M[G_0]$, $M[G_1]$, and $M[G_2]$, and any two of them will be mutually generic, so we will have $M[G_i]\cap M[G_j]=M$ for $i\neq j$. But once you have two of them, you can define the third, and so you get the whole thing.

Of course, this picture isn't the complete lattice of grounds, since every Cohen real can be split in two, and so if you had meant to ask about that one would need a more precise question.

Meanwhile, you ask about all possible $\mathbb{P}_i$, but the answer in this generality is false.

Here is one quick way to make an example. Take a Suslin tree $T$ that is $3$-fold Suslin off the generic branch. This is a strong form of rigidity, which means that after forcing with the tree, what remains above any node not on the generic branch is still Suslin. And again.

If we have any three generics for this forcing, then they will be automatically mutually generic. And so we can combine any two to form the model $M[G_0,G_1]$, which will not have $G_2$, and so it won't conform with your lattic.