The triangle $ ABC $ is inscribed in the circle $ \omega $. Points $ K, L, M $ are marked on the sides $ AB, BC, CA $, respectively, and $ CM \cdot CL = AM \cdot BL $. Ray $ LK $ intersects line $ AC $ at point $ P $. The common chord of the circle $ \omega $ and the circumscribed circle of the triangle $ KMP $ meets the segment $ AM $ at the point $ S $. Prove that $ SK \parallel BC $.
