In the quadrilateral $ABCD $, $AB = BC $, the point $K $ is the midpoint of the side $CD $, the rays $BK $ and $AD $ intersect at the point $M $ , the circumscribed circle $ \Delta ABM $ intersects the line $AC $ for the second time at the point $P $. Prove that $\angle BKP = 90 {} ^ \circ $. (Anton Trygub)