58th Ukrainian National Mathematical Olympiad

Let $T$ be the point, symmetrical to $M$ with respect to $AK$ (Fig. 4). Obviously, to prove the statement, it suffices to prove that $BT\parallel AK$. Notice that points $C,P,T$ lie on the same line. Also, $$\angle TAB=\angle TAK-\angle BAK=\angle MAK-\angle KAC=\angle CAM=\angle TCB,$$ Which means the quadrilateral $BCAT$ is inscribed, and $$\angle TBA=\angle TCA=\angle PAC=\angle PAB,$$ which proves the parallelism.