2014/15 AHSMC Part 1 Problem 14

Note that $AC \le AP + PC = 14$, $BD \le BP + PD = 24$ and the area of $ABCD$ is at most $\dfrac{1}{2} AC \cdot BD \le 168$. Thus we must have equality, so that $P$ is the point of intersection of the diagonals which must be perpendicular to each other. By Pythagoras' Theorem, $AB = AD = \sqrt{9^2 + 12^2} = 15$ and $CB = CD = \sqrt{5^2 + 12^2} = 13$. Hence the perimeter of $ABCD$ is $2(15 + 13) = 56$. The answer is (B).