Let $B'$, $C'$ be the intersections of the perpendicular bisector of $HD$ with $BH$, $CH$ respectively.

Denote by $M$, $M'$, $M_0$ the midpoints of $BC$, $B'C'$, $EF$, respectively. Since $N$ is the center of the circle $FEC'B'$, we have
$$\angle NM_0P = \angle NM'P = 90^\circ$$
Let the intersections of $HM$ with the circumcircle of $ABC$ be $K$, $A'$ (where $A$, $A'$ are on opposite sides of $BC$). We know that $\angle AKM = 90^\circ$ and quadrilateral $AEFK$ is cyclic. We have
$$\begin{aligned} \angle FKE &= \angle FAE \\ \angle KEF &= \angle KAF = \angle KCB \end{aligned}$$
So triangles $KBC$, $KFE$ are similar, and consequently, we have $\angle KM_0F = \angle KMB = \angle KM'B'$, resulting $KM_0M'P$ is cyclic.

Claim 2. If $A_0$ is the intersection of the line through $A$ parallel to $BC$ with the circumcircle, then $N$, $L$, $A_0$ are collinear.
Proof. Let $O$ be the circumcenter of $ABC$, and let the intersections of $OM$ with the circumcircle be $X$, $Y$ (where $X$ is on the arc of $BC$ that contains $A$). We know $N$ is the midpoint of $OH$, thus
$$\angle NM'K = \angle OMK.$$
On the other side,
$$\begin{aligned} \frac{\widehat{XK}}{2} &= \frac{\widehat{AK}}{2} + \frac{\angle B - \angle C}{2} \\ \frac{\widehat{YA'}}{2} &= \frac{\angle B - \angle C}{2} \\ \angle OMK &= \frac{\widehat{YA'}}{2} + \frac{\widehat{XK}}{2} = \frac{\widehat{AK}}{2} + (\angle B - \angle C) \end{aligned}$$

Also, since $\frac{AA_0}{2} = \angle B - \angle C$ and $\angle KLN = \angle OMK$, then $L$, $N$, $A_0$ are collinear.
Let $H'$ be the reflection of $H$ with respect to $BC$, it follows that this point is on the circumcircle and $A_0$, $O$, $H'$ are collinear. We have $\angle ADN = \angle AH'A_0 = \angle ALA_0$, which means $ANDL$ is indeed cyclic and the statement is proved.