Let $ E\^{D}N=N\^{D}C=x$, $ D\^{E}N=N\^{E}B=y$ and $ P\^{D}E=P\^{E}D=z$. Then, we have $ M\^{D}P=y$ and $ M\^{E}P=x$. Let $ K(\neq E)$, $ L(\neq D)$ the intersection points of the circuncircle of the triangle $ MDE$ with the lines $ EP$, $ DL$. We have: $ M\^{K}E=M\^{D}E=y+z=K\^{E}N$, so $ MK$ and $ EN$ are parallels. $ M\^{L}D=M\^{E}D=x+z=L\^{D}N$, so $ LM$ and $ ND$ are parallels. $ L\^{K}E=L\^{D}E=z=K\^{E}D$, so $ LK$ and $ ED$ are parallels. Then, there exists and homotety which transforms the triangle $ LMK$ into the triangle $ DNE$. The centre of that homotety is $ KE\cap LD=\{P\}$. In particular, $ M$, $ P$ and $ N$ are collinears.