Triangle $PAB$ and square $ABCD$ are in perpendicular planes. Given that $PA=3$, $PB=4$, and $AB=5$, what is $PD$? [asy]
size(100); defaultpen(linewidth(0.8)); dotfactor=4;
draw((0,-0.5)--(0,3)--(5,3)--(5,-0.5)--cycle);
filldraw((0,0)--(-2,-2)--(3,-2)--(5,0)--cycle,white,defaultpen);
draw((1.5,0)--(3.5,0)--(3.5,2)--(1.5,2)--cycle);
draw((1.5,0)--(1.6,-0.5)--(3.5,0));
dot("$B$",(1.5,0),SW); dot("$A$",(3.5,0),SE); dot("$C$",(1.5,2),W); dot("$D$",(3.5,2),E); dot("$P$",(1.6,-0.5),S);
[/asy]
Since line segment $AD$ is perpendicular to the plane of $PAB$, angle $PAD$ is a right angle. In right triangle $PAD, PA=3 \mbox{ and }AD=AB=5$. By the Pythagorean Theorem $PD = \sqrt{3^2+5^2}=\boxed{\sqrt{34}}$. The fact that $PB=4$ was not needed.