$A+P=(a+2)(b+2)-4=N$ is achievable iff $N+4$ has a factorization into two factors both $\ge3$.
$158+4=162=2\cdot3^4$: factor pairs with both $\ge3$ include $(3,54),(6,27),(9,18)$ -- achievable. $156+4=160=2^5\cdot5$: $(4,40),(5,32),(8,20),(10,16)$ -- achievable.
$154+4=158=2\cdot79$ with $79$ prime: only factor pairs are $(1,158),(2,79)$, neither with both factors $\ge3$ -- \emph{not} achievable.