Write the side of the square as $x$ , and the diagonals as $\sqrt{2}x$ Since $P$ is on the circumcircle of $ABCD$ the quadrilaterals $APBD$ and $APBC$ are cyclic and thus, applying Ptolemy's theorem on both we get : $PD=PA + \sqrt{2}BP$ $PC=\sqrt{2}AP+BP$ Hence, $PB+PC=(\sqrt{2}+1)(AP+BP)$ => $\frac{AP+BP}{CP+DP} = \frac{1}{\sqrt{2}+1}=\sqrt{2}-1$ A generilization of it was shortlisted for the Balkan MO 2012 https://artofproblemsolving.com/community/c6h1708935p11012178

In APCD, BY Ptolemy, PA+PC=PD√2 PA=PD√2-PC IN PBCD, BY PTOLEMY, PB+PD=PC√2 PB=PC√2-PD PA+PB=(PC+PD)(√2-1) (PA+PB)/(PC+PD)=√2-1