Let $\angle{PBD}=x$ and circumradius as $R$ so $PC=2RSin x,PB=2RSin (45^0-x),PA=2RCos x,PD=2RCos (45^0-x)$ So if all are rational then $\frac{PC}{PA}=tan x$ is rational but from $\frac {PB}{PC}$ we get t $tanx$ is irrational. Contra !!

note that , if $P$ coincides with any of $A,B,C,D$ , then all of them can't be rational , since in a square , $diagonal=\sqrt{2}side$ WLOG ,the circumcircle of $ABCD$ is $\mid z \mid=r$ assume that , $a=r$,$b=-r$ ,$c=ri$ ,$d=-ri$ and $P=re^{i\theta}$ if possible , assume that , all the distances are rational. $PA=2r\mid sin{\frac{\theta}{2}} \mid$. $PB=2r\mid cos{\frac{\theta}{2}}\mid$ $PC=2r\mid sin({\pi}/4-{\theta/2})\mid$ .$PD=2r \mid cos({\pi/4}-{\theta/2}) \mid$ note that , all these distances are $>0$. so , $rsin(\theta/2)$ and $rcos(\theta/2)$ are rational. now , its a trivial job to see that both $PC$ and $PD$ can't be rational.so , a contradiction!