• $AP^2+CP^2=DP^2+BP^2$
• $CP^2=AB^2=AP^2+BP^2$

Let $R$ and $T$ be on $BC$ and $DA$ respectively such that 

$PR\perp BC,~PT\perp DA$

So, $CR=DT$ and $AT=BR$

\[AP^2+CP^2=(AT^2+PT^2)+(PR^2+CR^2)=(BR^2+PR^2)+(PT^2+DT^2)=BP^2+DP^2\]

Now, $CP=CD=AB$. 

So,

$\implies 2(DP^2+PB^2)=DP^2+PB^2+AP^2+PC^2=PC^2+AB^2+DP^2$

$\implies DP^2=2(AB^2-PB^2)=2AP^2\implies DP=\sqrt{2}AP$

As wanted, $DP=\sqrt{2}AP$.