Let A-altitude cut circumcircle $(O)$ of $\triangle ABC$ again at $X$. Let $AX$ cut $\omega$ at $P, P'$. $PK$ cuts $(O)$ at $E, F$. Let $PQ$ be diameter of $\omega$. $P'Q \parallel BC$ (both $\perp$ $AX$) $\implies$ $OK$ is their common perpendicular bisector, cutting $\omega$ again at $O'$ $\implies$ $PO, PO'$ bisect $\measuredangle (APP', EPF)$ $\implies$ $AEXF$ is isosceles trapezoid inscribed in $(O)$, either convex or cross, with diagonals $AX, EF$ and diagonal intersecttion $P$ $\implies$ either $PE = PA$ or $PF = PA$, depending on $E, F$ notation.