The inscribed circle $\Omega$ of triangle $ABC$ touches the sides $AB$ and $AC$ at points $K$ and $ L$, respectively. The line $BL$ intersects the circle $\Omega$ for the second time at the point $M$. The circle $\omega$ passes through the point $M$ and is tangent to the lines $AB$ and $BC$ at the points $P$ and $Q$, respectively. Let $N$ be the second intersection point of circles $\omega$ and $\Omega$, which is different from $M$. Prove that if $KM \parallel AC$ then the points $P, N$ and $L$ lie on one line.