Let $M$ be the midpoint of the arc $BAC$ of the circumcircle of the triangle $ABC,$ $I$ the incenter of the triangle $ABC$ and $L$ a point on the side $BC$ such that $AL$ is bisector. The line $MI$ cuts the circumcircle again at $K.$ The circumcircle of the triangle $AKL$ cuts the line $BC$ again at $P.$ Prove that $\angle AIP = 90^{\circ}.$