South African Mathematics Olympiad

Since $AD=AE$, $AED$ is isosceles, so we also have $\angle ADE=\angle AED$. Combining this with the fact that $ABPE$ and $ABPF$ are cyclic, we obtain

$$\angle ABP=180^{\circ }-\angle AEP=\angle AED=\angle ADE=\angle ADP$$ as well as $$\angle ABP=180^{\circ }-\angle AFP=\angle DFP,$$ so $\angle FDP=\angle DFP$, which implies that $PD=PF$. Likewise, $\angle ABE=\angle AEB$ since $AB=AE$, which means that $$\angle APB=\angle AEB=\angle ABE=\angle APE=\angle APD.$$ Now we see that triangles $ABP$ and $ADP$ have the same angles ($\angle APB=\angle APD$, $\angle ABP=\angle ADP$) and the common side $AP$, so they are congruent. Thus $PB=PD=PF$, which means that $P$ is the circumcentre of $BDF$.