We will make several uses of the fact that the lines connecting a vertex of a triangle with the circumcentre and the orthocentre are symmetrical with respect to the angle bisector at this vertex.
Let $AC > BC$, $AA_1$ and $BB_1$ be the altitudes of the triangle $ABC$, $O$ its circumcentre. The points $A_1$, $B_1$ and $R$ lie on the circle with diameter $CH$. The triangles $A_1B_1C$ and $ABC$ are similar, and their similarity ratio is $\frac{A_1C}{AC} = \cos C$. Then the ratio of the circumradii of these triangles is also $\cos C$, that is, $\frac{CN}{AO} = \cos C$, hence $CN = AO \cos C = OM$. Since $OM \parallel CN$, $COMN$ is a parallelogram. Let $\angle ACO = \angle BCH = x$, $\angle OCM = \angle NMC = y$, $\angle RCT = \angle RNT = z$. Then, since $NC = NR$, we have $z + \angle TCN = \angle RCN = \angle CRN = y + z$, whence, $\angle TCN = y$ and the lines $CT$ and $CM$ are symmetrical with respect to the bisector of $ACB$. Since $\angle AHB_1 = \angle ACB$ and $\angle AHP = RHB_1 = \angle B_1CR = \angle BCT$, we have $\angle RCT = \angle RHP$, that is, $C$, $R$, $Q$ and $H$ belong to the same circle with diameter $CH$ (here $Q = CT \cap PH$). The desired perpendicularity follows immediately.

Note. We have proved in this solution that $CT$ lies on the symmedian of the triangle $ABC$.

Let $AA_1$, $BB_1$, and $CC_1$ be the altitudes of the triangle $ABC$, and $T_1$ the midpoint of $A_1B_1$. Then $MA_1 = MB_1 = AB/2$, $NA_1 = NB_1 = CH/2$, therefore $MN$ is the perpendicular bisector of $A_1B_1$. We will prove that $T_1 = T$.
Since $CM$ and $CT_1$ are respective medians in similar triangles $ABC$ and $A_1B_1C$, while $AB$ and $A_1B_1$ are anti-parallel, $CT_1$ lies on the symmedian $CS$ of the triangle $ABC$. Let $\angle NRC = \angle RCN = \delta$. Then $90^\circ - \delta = \angle CMB = \angle CT_1B_1 = \angle ST_1A_1$, hence $\delta = \angle MT_1S = \angle CT_1N$, that is, $\angle NRC = \angle CT_1N$, and it follows that $T_1$ belongs to the circumcircle of the triangle $CNR$.
Next, as in the first solution, we can prove $HP \perp CT$. The proof is indeed complete.