Three collinear points

[ins-]

Let $ABCD$ is an inscribed quadrilateral in a circle $k$. $P$ is the intersection point of the diagonals. $E$ is the intersection point of $AB$ and $CD$. $F$ is the intersection point of $AD$ and $BC$. $k_1$ is the circumcircle of the triangle $AEF$. $k_2$ is the circumcircle of the triangle $CEF$. $M$ is the intersection point of $k$ and $k_1$. $N$ is the intersection point of $k$ and $k_2$. Prove that $M$, $N$ and $P$ are collinear.

[Dimoniada]

Solution in outline:
Note, that $FE$ - polar of point $P$ relative to $k$.
Radical axis $AM, CN, FE$ of $(k\cap k_1), (k\cap k_2), (k_1\cap k_2)$ intersects at common point $L$, lying on $FE$.
Because $(AM\cap CN) \in FE$ and $[APC]$ - one line, then $[MPN]$ - line too.

[ins-]

I suppose this problem can be solved without using polar and probably radical axis, but it will be harder.