Answer: E

Small Hint:

For a quadrilateral with diagonals $d_1,d_2,$ the area is at most $\tfrac12 d_1 d_2,$ with equality when the diagonals are perpendicular

Big Hint:

Here $\tfrac12(PA+PC)(PB+PD)$ $=2002$ forces perpendicular diagonals meeting at $P$; find each side with the Pythagorean theorem

Solution:

For any quadrilateral, the area is at most $\tfrac12\,d_1 d_2$ where $d_1,d_2$ are the diagonals, with equality exactly when they are perpendicular. Here $$\begin{gathered} 2002=\text{Area} \\ {}\le \tfrac12\,AC\cdot BD \\ {}\le \tfrac12(PA+PC)(PB+PD) \\ {}= \tfrac12\cdot52\cdot77 \\ {}= 2002. \end{gathered}$$

Equality forces the diagonals to be perpendicular and to intersect at $P.$ Then $$\begin{aligned} AB &= \sqrt{24^2+32^2}=40, \\ BC &= \sqrt{28^2+32^2}=4\sqrt{113}, \end{aligned}$$ $$\begin{aligned} CD &= \sqrt{28^2+45^2}=53, \\ DA &= \sqrt{45^2+24^2}=51. \end{aligned}$$

The perimeter is $144+4\sqrt{113}=4\left(36+\sqrt{113}\right).$

Thus, the correct answer is E.