First notice that $\angle PBM=90^{\circ}-\angle BAM$ and $\angle QCM=90^{\circ}-\angle CAM$, then $\angle BLC=180^{\circ}-\angle PBM-\angle CAM=\angle BAM+\angle CAM=\angle BAC$. Since K is the reflection of L across $PQ$ we have $\angle BLC=\angle PKQ$. By radical axis note that $PQ$ is the perpendicular bisector of $AM$ with this and knowing $P, Q$ are centers of $\triangle ABM$, $\triangle ACM$, we have $\angle APQ=\angle MPQ=\angle ABM$, and $\angle AQP=\angle MQP=\angle ACM$, so $\angle PMQ=180^{\circ}-\angle MPQ-\angle MQP=180-\angle ABM-\angle ACM=\angle BAC=\angle PKQ$ and the result follows