Nathaniel Johnston's comment is correct; this is indeed a variant of the matrix tree theorem. To see this, one needs to rename the variable $P_{i,i}$ to $P_{i,0}$, for each $i$.

Then, $A$ becomes the matrix obtained by removing the $0$th row and column from the $(n+1)\times (n+1)$ matrix $$ L = \begin{bmatrix} * & -P_{0,1} & -P_{02} & -P_{0,3}\\ -P_{1,0} & * & -P_{1,2} & -P_{1,3}\\ -P_{2,0} & -P_{2,1} & * & -P_{2,3}\\ -P_{3,0} & -P_{3,1} & -P_{3,2} & * \end{bmatrix}, $$ where the diagonal entries denoted by $*$ are chosen so that each row sums to $0$. This matrix is the Laplacian of a weighted directed graph, and we can apply the generalized matrix tree theorem suggested in the comment (the first unnumbered theorem on page 379 of https://doi.org/10.1016/0097-3165(78)90067-5). This theorem states that $\det A$ is the sum of the weights of all arborescences (directed trees) rooted in the vertex $v_0$, each of which is a product of $n$ variables of the form $P_{ij}$.