Cute Extend $AE$ to cut the circumcircle at $S$. $OP=OQ$ is equivalent to $SQ=AP$ (pop) which is easy by congruent triangles with some angles.

Let $BP$ intersect the circumcircle again at $X$, and let $CQ$ intersect the circumcircle again at $Y$. By Reim's theorem/homothety, we have that $BY\parallel AE\parallel CX$. In particular, $BYXC$ is an isosceles trapezoid with $BC=XY$, and its diagonals are symmetric across the perpendicular bisectors of $BY$ and $CX$, so $OP=OQ$, as desired.