We note the half of the angle at $A$ as $\alpha = \frac{1}{2} \angle BAC = \angle BAM = \angle MAC$, as $M$ is the midpoint of arc $BC$. Then we note that $\angle ABP = \angle ABD = 90^\circ - \angle DAB = 90^\circ - \alpha$ and that $\angle EDA = 90^\circ - \angle DAE = 90^\circ - \alpha$. Moreover, because of the straight angle $\angle ADM$, we find that $\angle PDE = 180^\circ - \angle EDA - \angle MDP = 180^\circ - (90^\circ - \alpha) - 90^\circ = \alpha$.

Now we define $K$ as the second intersection of the circumscribed circle of $\triangle ADE$ with $\Gamma$ (in addition to $A$). Then we note that $\angle AKE = \angle ADE = 90^\circ - \alpha = \angle ABP = \angle AKP$, so $K$, $E$ and $P$ are collinear. Similarly, $\angle AKD = 180^\circ - \angle AED = 90^\circ = \angle AKN$, due to Thales because $A$ and $N$ are antipodes. So $K$, $D$ and $N$ are collinear.

For the last line, we claim that $BFDK$ is also a cyclic quadrilateral. Indeed, $\angle KBF = \angle KBC = 180^\circ - \angle KAC = 180^\circ - \angle KAE = \angle KDE = 180^\circ - \angle KDF$. (Note that this is in fact Miquel's theorem in $\triangle CEF$ and cyclic quadrilateral $AKBC$ and $DEAK$.) From the inscribed angle theorem in cyclic quadrilateral $BFDK$, it follows that $\angle BKF = \angle BDF = \angle PDE = \alpha = \angle BAM = \angle BKM$, so $K$, $F$ and $M$ are collinear. We conclude that $ND$, $MF$ and $PE$ are concurrent in vertex $K$. $\square$