Because $APEF$ is cyclic and $D=AF \cap EP$, the triangles $DFP$ and $DAE$ are similar. Since $AD=AE$, this implies $PD=PF$. Moreover, because $A$ is the circumcenter of $\triangle BED$, we have $\angle DEB = 180^\circ-\frac{\angle BAD}{2}$ and hence $\angle BAP = \angle BEP = \angle CED = \frac{\angle BAD}{2} = \frac{\angle BAF}{2}$. Since $P$ is on the circumcircle of $\triangle ABF$, this implies $PB=PF$. Hence the result.