smc

Place the square in the $xy$-plane with $A$ as the origin, so that $B=(1,0),C=(1,1),$ and $D=(0,1).$ We are given that $P{A}^2+P{B}^2=P{C}^2,$ so $$\begin{array}{rl}& (x^2+y^2)+((x-1{)}^2+y^2)=(x-1{)}^2+(y-1{)}^2\\ \Rightarrow & 2x^2+2y^2-2x+1=x^2+y^2-2x-2y+2\\ \Rightarrow & x^2+y^2=-2y+1\\ \Rightarrow & x^2+y^2+2y-1=0\\ \Rightarrow & x^2+(y+1{)}^2=2.\end{array}$$ Thus we see that $P$ lies on a circle centered at $(0,-1)$ with radius $\sqrt{2}.$ The farthest point from $D$ on this circle is at the bottom of the circle, at $\left(0,-1-\sqrt{2}\right),$ in which case $PD$ is $1-\left(-1-\sqrt{2}\right)=\boxed{2+\sqrt{2}}.$