$O$ is the center of $\omega$ and $M \equiv AC \cap OL,$ $N \equiv AB \cap OK$ are the midpoints of $AC,AB.$ Since $AB$ is the polar of $K$ with respect to $\omega,$ then $PK \perp ON$ is the polar of $N$ with respect to $\omega.$ Likewise, $PL$ is the polar of $M$ with respect to $\omega.$ Therefore, $P$ is the pole of the A-midline $MN$ with respect to $\omega$ $\Longrightarrow$ $P$ lies on the perpendicular from $O$ to $MN,$ i.e. the perpendicular bisector of $\overline{BC}$ $\Longrightarrow$ $PB=PC.$