Saudi Arabian Mathematical Competitions

Suppose that $KA$ cuts $PF$ at $M$, $KB$ cuts $PG$ at $N$. By angle chasing, we have
$$\angle IEF = \angle BAI = \angle FAI,$$
then $AIFE$ is cyclic. In addition, we get
$$\begin{aligned} \angle AMF = \angle KAI & = \angle KAB + \angle BAI \\ & = \angle AEI + \angle FEI = \angle AEF \end{aligned}$$
then $AIFM$ is cyclic.

Summarily, points $A, I, F, E, M$ lie on a circle. And similarly, points $B, I, G, D, N$ also lie on a circle. We have known that $DE$ is perpendicular bisector of $CI$, so it is easy to see that $IFCG$ is a rhombus. From that above, we get
$$\angle AMI = \angle AEI = \angle KAB,$$
then $IM \parallel AB$. Similarly $IN \parallel AB$, implies that $M, N, I$ are collinear. But from $PF\parallel AD$, $PG\parallel BE$, we have $AIFM$ and $BIGN$ are isosceles trapezoids, so
$$\begin{equation*} AM = IF = IG = BN, \tag{1} \end{equation*}$$
then $ABNM$ is also an isosceles trapezoid, which means that it is cyclic.

In the other hand, we get
$$\begin{equation*} \angle PFG = \angle MFE = \angle MIE = \angle BIN = \angle ING, \tag{2} \end{equation*}$$
so $FGNM$ is cyclic.

From (1) and (2), we have $PK$ is the radical axis of two circles $(AIE)$ and $(BID)$. Now on, we can easily take the conclusion of this problem. $\square$