Complete quadrilateral

[ins-]

Let $ABCD$ be quadrilateral inscribed in a circle $k_1$. Let $E$ be the common point of the lines $AB$ and $CD$, $F$ be the common point of the lines $AD$ and $BC$. Let $k_2$ be the circle circumscribing the triangle $BEF$, and $P$ be the intersection point of $k_1$ and $k_2$, different from $B$. Prove that the line $P$, $D$ and the middle of $EF$ lie on the same straight line.

I'm sorry for my bad English

[ins-]

\angle CDP=180 -\angle CBP=\angle PEF=\angle EDM \leftrightarrow EM^2=ED.EP
\angle ABP=\angle EFP =\angle ADP=\angle MDF \leftrightarrow FM^2=ED.EP
It is the solution of my problem posted by user ubuntu from math10.com/bg

I would like ask you
1. Do you like the problem?
2. What is its level of difficulty?
3. Is it well known fact?