Let $ABC$ be an acute triangle (with $AB \ne AC$) and $M$ be the midpoint of $BC$. The circle of diameter $AM$ cuts $AC$ at $N$ and $BC$ again at $H$. A point $K$ is taken on $AC$ (between $A$ and $N$) such that $CN = NK$. Segments $AH$ and $BK$ intersect at $L$. The circle that passes through $A$,$K$ and $L$ cuts $AB$ at $P$. Prove that $C$,$L$ and $P$ are collinear.