We will prove that the fixed point is the orthocenter $H$ of triangle $ABC$. Let $X$ be the foot of the perpendicular from $C$ to $AB$, and let $Y$ be the foot of the perpendicular from $B$ to $AC$. Note that if $M$ is the midpoint of $BC$, then $MB = MX = MY = MC$. Suppose without loss of generality that $P$ is between $B$ and $M$. Then $D$ is between $B$ and $X$, and $E$ is between $A$ and $Y$. The quadrilateral $AXHY$ is cyclic, since $\angle AXH = \angle AYH = 90^\circ$. To show that $ADHE$ is also cyclic, it suffices to show that $\triangle DHX \sim \triangle EHY$, or that
$$\frac{DX}{EY} = \frac{XH}{YH}. \qquad (38)$$
Applying the Law of Sines in the cyclic quadrilateral $AXHY$, we find that $\frac{XH}{YH} = \frac{\cos B}{\cos C}$. Note that $DX = BX - BD = BC \cos B - 2BP \cos B$. Similarly, $EY = EC - CY = 2PC \cos C - BC \cos C$. Hence, we find
$$\frac{DX}{EY} = \frac{BC - 2BP}{2PC - BC} \cdot \frac{\cos B}{\cos C} = \frac{\cos B}{\cos C}.$$

We adopt the notations of the previous solution and operate on the same assumptions on the configuration. We give an alternate way to see (38). Let $U_B$ and $U_C$ be the foot of the perpendicular from $M$ to $AB$ and $AC$, respectively. Let $V_B$ and $V_C$ be the feet of the perpendiculars from $P$ to $AB$ and $AC$, respectively. Note that
$$\frac{DX}{EY} = \frac{2U_BV_B}{2U_CV_C} = \frac{U_BV_B}{U_CV_C}.$$
Let $S$ be the perpendicular of $P$ to $U_BM$, and let $T$ be the foot of the perpendicular from $M$ to $PV_C$. Suppose that $PS$ and $MT$ intersect at $R$. Note that $\triangle RPM \sim \triangle ABC$. Points $S$ and $T$ in $RPM$ correspond to points $X$ and $Y$ in $ABC$. Hence, we see that
$$\frac{U_B V_B}{U_C V_C} = \frac{PS}{TM} = \frac{BX}{CY} = \frac{HX}{HY}.$$