China Hong Kong Mathematical Olympiad

We only work on the configuration as shown since the other cases are similar.
We have
$$\angle CPA = \angle BAP + \angle CBA + \angle PCB = 90^\circ + \angle CBA.$$
Let $O_1$ be the centre of $(APC)$. Then we find that
$$\angle AO_1C = 360^\circ - 2\angle CPA = 180^\circ - 2\angle CBA = 180^\circ - 2\angle COA.$$
This shows $A$, $O$, $C$, $O_1$ are concyclic. As $O$, $O_1$ lie on the perpendicular bisector of $AC$, $O_1$ is just the intersection of the tangents at $A$ and $C$ to $(ABCD)$. Similarly, let $O_2$ be the intersection of the tangents at $B$ and $D$ to $(ABCD)$. Then $O_2$ is the centre of $(BPD)$.

Clearly, $P$ lies on the radical axis of $(APC)$ and $(BPD)$. Secondly, since we have $EA \times EC = EB \times ED$, $E$ also lies on this radical axis. Lastly, the powers of $O$ with respect to the two circles are $OA^2$ and $OB^2$ respectively as $OA$ and $OB$ are the tangents. Since $OA = OB$, these powers are equal so that $O$ lies on the radical axis. Therefore, $P$, $E$, $O$ are collinear.