证明：因为四边形$ABCD$是正方形，所以$BA = BC，$$\angle ABC = 90^{\circ}，$所以$\angle ABP + \angle CBP = 90^{\circ}。$
因为$BE \perp PB，$所以$\angle PBE = 90^{\circ}，$所以$\angle ABP + \angle ABE = 90^{\circ}，$所以$\angle ABE = \angle CBP。$
因为四边形$ABCP$是圆的内接四边形，所以$\angle BAP + \angle BCP = 180^{\circ}。$
因为$\angle BAP + \angle BAE = 180^{\circ}，$所以$\angle BAE = \angle BCP。$
在$\triangle ABE$和$\triangle CBP$中，
$\begin{cases}\angle ABE = \angle CBP \\BA = BC \\\angle BAE = \angle BCP\end{cases}$
所以$\triangle ABE \cong \triangle CBP，$所以$EA = PC，$$BE = BP，$所以$PA + PC = PA + EA = PE=\sqrt{BP^{2}+BE^{2}}=\sqrt{2}BE。$
