AHSME 1996 Problem 9

Triangle $PAB$ and square $ABCD$ are in perpendicular planes. Given that $PA = 3, PB = 4$ and $AB = 5$, what is $PD$?

$\text{(A)}\ 5 \text{(B)}\ \sqrt{34} \text{(C)}\ \sqrt{41} \text{(D)}\ 2\sqrt{13} \text{(E)}\ 8$