The answer to your first question is yes. Let $C \subseteq B$ be as in your remark. Let $\pi : B \to C$ be the standard projection map, $\pi(b) = \inf \{ c \in C : c \geq b \}$. The restriction of $\pi$ to $P$ maps a dense subset of $B$ onto a dense subset $Q \subseteq C$, it is a projection, and $|Q| \leq |P|$.
For the second question, the answer is no. For example, it is well-known that there is a dense embedding from $\mathrm{Col}(\omega,\omega_1)$ into $\mathrm{Add}(\omega,\omega_1) \times \mathrm{Col}(\omega,\omega_1)$. It follows that there is a projection from $\mathcal{B}(\mathrm{Col}(\omega,\omega_1))$ to $\mathcal{B}(\mathrm{Add}(\omega,\omega_1))$. But there is no map $\pi$ from the raw partial orders with your desired properties.
Suppose that such a $\pi$ existed, let $G$ be $\mathrm{Col}(\omega,\omega_1)$-generic, and let $H = \pi[G]$. Then $\omega_1$ is preserved in $V[H]$, and $\pi^{-1}[H]$ is uncountable. By pigeonhole, there are $p,q \in \pi^{-1}[H]$ such that for some $n$, $p(n) \not= q(n)$. This contradicts the requirement that $G$ is a filter.
