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 $\triangle ABC$ and let $O$ be circumcenter of $\triangle ABC$. 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.