Solution:

Let the side lengths be positive integers $x$ and $y$. Then $$A+P=xy+2x+2y=(x+2)(y+2)-4.$$ Hence $A+P+4$ must factor into two integers both at least $3$.

The answer choices plus $4$ are $104,106,108,110,112$. All except $106$ have a factorization with both factors at least $3$: $$104=4\cdot26, \quad 108=9\cdot12, \quad 110=10\cdot11, \quad 112=7\cdot16.$$ But $106=2\cdot53$, so it cannot equal $(x+2)(y+2)$.

Thus, B is the correct answer.