We keep triangle $ABP$ fixed and move the line $CD$ parallel to itself uniformly, i.e. linearly dependent on a single parameter $\lambda$ (see Figure 1). Then the points $C$ and $D$ also move uniformly. Hence, the points $O_{2}, H_{2}$ and $E_{2}$ move uniformly, too. Therefore also the perpendicular from $E_{2}$ on $AB$ moves uniformly. Obviously, the points $O_{1}, H_{1}, E_{1}$ and the perpendicular from $E_{1}$ on $CD$ do not move at all. Hence, the intersection point $S$ of these two perpendiculars moves uniformly. Since $H_{1}$ does not move, while $H_{2}$ and $S$ move uniformly along parallel lines (both are perpendicular to $CD$), it is sufficient to prove their collinearity for two different positions of $CD$.

Figure 1

Let $CD$ pass through either point $A$ or point $B$. Note that by hypothesis these two cases are different. We will consider the case $A \in CD$, i.e. $A = D$. So we have to show that the perpendiculars from $E_{1}$ on $AC$ and from $E_{2}$ on $AB$ intersect on the altitude $AH$ of triangle $ABC$ (see Figure 2).

Figure 2

To this end, we consider the midpoints $A_{1}, B_{1}, C_{1}$ of $BC, CA, AB$, respectively. As $E_{1}$ is the center of FEUERBACH's circle (nine-point circle) of $\triangle ABP$, we have $E_{1}C_{1} = E_{1}H$. Similarly, $E_{2}B_{1} = E_{2}H$. Note further that a point $X$ lies on the perpendicular from $E_{1}$ on $A_{1}C_{1}$ if and only if
$$XC_{1}^{2} - XA_{1}^{2} = E_{1}C_{1}^{2} - E_{1}A_{1}^{2}.$$
Similarly, the perpendicular from $E_{2}$ on $A_{1}B_{1}$ is characterized by
$$XA_{1}^{2} - XB_{1}^{2} = E_{2}A_{1}^{2} - E_{2}B_{1}^{2}$$
The line $H_{1}H_{2}$, which is perpendicular to $B_{1}C_{1}$ and contains $A$, is given by
$$XB_{1}^{2} - XC_{1}^{2} = AB_{1}^{2} - AC_{1}^{2}.$$
The three lines are concurrent if and only if
$$\begin{aligned} 0 & = XC_{1}^{2} - XA_{1}^{2} + XA_{1}^{2} - XB_{1}^{2} + XB_{1}^{2} - XC_{1}^{2} \\ & = E_{1}C_{1}^{2} - E_{1}A_{1}^{2} + E_{2}A_{1}^{2} - E_{2}B_{1}^{2} + AB_{1}^{2} - AC_{1}^{2} \\ & = -E_{1}A_{1}^{2} + E_{2}A_{1}^{2} + E_{1}H^{2} - E_{2}H^{2} + AB_{1}^{2} - AC_{1}^{2} \end{aligned}$$
i.e. it suffices to show that
$$E_{1}A_{1}^{2} - E_{2}A_{1}^{2} - E_{1}H^{2} + E_{2}H^{2} = \frac{AC^{2} - AB^{2}}{4}$$
We have
$$\frac{AC^{2} - AB^{2}}{4} = \frac{HC^{2} - HB^{2}}{4} = \frac{(HC + HB)(HC - HB)}{4} = \frac{HA_{1} \cdot BC}{2}.$$
Let $F_{1}, F_{2}$ be the projections of $E_{1}, E_{2}$ on $BC$. Obviously, these are the midpoints of $HP_{1}$, $HP_{2}$, where $P_{1}, P_{2}$ are the midpoints of $PB$ and $PC$ respectively. Then
$$\begin{aligned} & E_{1}A_{1}^{2} - E_{2}A_{1}^{2} - E_{1}H^{2} + E_{2}H^{2} \\ & = F_{1}A_{1}^{2} - F_{1}H^{2} - F_{2}A_{1}^{2} + F_{2}H^{2} \\ & = \left(F_{1}A_{1} - F_{1}H\right)\left(F_{1}A_{1} + F_{1}H\right) - \left(F_{2}A_{1} - F_{2}H\right)\left(F_{2}A_{1} + F_{2}H\right) \\ & = A_{1}H \cdot \left(A_{1}P_{1} - A_{1}P_{2}\right) \\ & = \frac{A_{1}H \cdot BC}{2} \\ & = \frac{AC^{2} - AB^{2}}{4}, \end{aligned}$$
which proves the claim.

Let the perpendicular from $E_{1}$ on $CD$ meet $PH_{1}$ at $X$, and the perpendicular from $E_{2}$ on $AB$ meet $PH_{2}$ at $Y$ (see Figure 3). Let $\varphi$ be the intersection angle of $AB$ and $CD$. Denote by $M, N$ the midpoints of $PH_{1}, PH_{2}$ respectively.

Figure 3

We will prove now that triangles $E_{1}XM$ and $E_{2}YN$ have equal angles at $E_{1}, E_{2}$, and supplementary angles at $X, Y$.
In the following, angles are understood as oriented, and equalities of angles modulo $180^{\circ}$.
Let $\alpha = \angle H_{2}PD$, $\psi = \angle DPC$, $\beta = \angle CPH_{1}$. Then $\alpha + \psi + \beta = \varphi$, $\angle E_{1}XH_{1} = \angle H_{2}YE_{2} = \varphi$, thus $\angle MXE_{1} + \angle NYE_{2} = 180^{\circ}$.
By considering the Feuerbach circle of $\triangle ABP$ whose center is $E_{1}$ and which goes through $M$, we have $\angle E_{1}MH_{1} = \psi + 2\beta$. Analogous considerations with the Feuerbach circle of $\triangle DCP$ yield $\angle H_{2}NE_{2} = \psi + 2\alpha$. Hence indeed $\angle XE_{1}M = \varphi - (\psi + 2\beta) = (\psi + 2\alpha) - \varphi = \angle YE_{2}N$. It follows now that
$$\frac{XM}{ME_{1}} = \frac{YN}{NE_{2}}.$$
Furthermore, $ME_{1}$ is half the circumradius of $\triangle ABP$, while $PH_{1}$ is the distance of $P$ to the orthocenter of that triangle, which is twice the circumradius times the cosine of $\psi$. Together with analogous reasoning for $\triangle DCP$ we have
$$\frac{ME_{1}}{PH_{1}} = \frac{1}{4 \cos \psi} = \frac{NE_{2}}{PH_{2}}.$$
By multiplication,
$$\frac{XM}{PH_{1}} = \frac{YN}{PH_{2}},$$
and therefore
$$\frac{PX}{XH_{1}} = \frac{H_{2}Y}{YP}.$$
Let $E_{1}X, E_{2}Y$ meet $H_{1}H_{2}$ in $R, S$ respectively.
Applying the intercept theorem to the parallels $E_{1}X, PH_{2}$ and center $H_{1}$ gives
$$\frac{H_{2}R}{RH_{1}} = \frac{PX}{XH_{1}},$$
while with parallels $E_{2}Y, PH_{1}$ and center $H_{2}$ we obtain
$$\frac{H_{2}S}{SH_{1}} = \frac{H_{2}Y}{YP}$$
Combination of the last three equalities yields that $R$ and $S$ coincide.