37th Iranian Mathematical Olympiad

Without loss of generality, let's assume that $AC\ge AB$. Let $D$ be the second intersection point of line $AH$ and the circumcircle of $△ABC$ and let $O$ be circumcenter of $△ABC$. Let the lines $OD$ and $BC$ meet at $J$. Then we have $\angle JDH=\angle JHD$ since $BC$ is perpendicular bisector of $HD$. On the other hand $\angle DCA=90-\angle B+\angle C$ so we have $\angle ODH=\angle B-\angle C$. Therefore $JH\parallel AO$, since $AO\perp EF$ so we get that $JH\perp EF$ and $J$ lies on the line $KH$. Note that $$\angle PKD=\angle AKD=90^{\circ }-\angle B+\angle C.$$ Also $$\angle OJP=\angle OJC=\angle CDJ+\angle DCJ=\angle C+90^{\circ }-\angle B.$$ Therefore $\angle OJP=\angle PKD$ and $JDKP$ is cyclic and $\angle PKH=\angle PKJ=\angle PDJ$. But we have $\angle PDJ=\angle PHJ=\angle PHK$ since $D$ is the reflection of $H$ with respect to $PJ$. So $\angle PKH=\angle PHK$ which gives us $PK=PH$ as desired.