Sorry for the lack of a diagram.

Let $HP$ intersect $AB$ at $M$. Since $\angle CHA = \angle CPA = 90$, $CHPA$ is cyclic, so $\angle HPC= \angle HAC = \angle OAB$ (well-known). Thus, $OPMA$ is cyclic, so $\angle OMA = \angle OPA = 90$ , so since the circumcenter $O$ lies on the perpendicular bisector of $AB$, we must have $AM=MB$.

Let the perpendicular bisector of $AB$ meet $BC$ at $Q$. Then $A,O,P,Q$ are concyclic, therefore $HP$, the Simson-line of $A$ w.r.t. the triangle $OCQ$, passes through the midpoint of $AB$.

Let the unit circle be $(ABC)$, and $c=1$. Then $m=\frac{1}{2} (a+b)$, $h=\frac{1}{2} (a+b+1- \frac{b}{a})$. Since, $P$, $O$, $C$ collinear, $p=k$ for $k \in \mathbb{R}$ and since $AP \perp OC$, \[ \frac{1}{k-a}+ \frac{1}{k-\frac{1}{a}}=0 \implies p=\frac{1}{2} (a+\frac{1}{a}) \]Consider the transformation $x \rightarrow 2x-a$. Then it suffices to show $b$, $b+1- \frac{b}{a}$, $\frac{1}{a}$ are collinear. \[ \frac{\frac{b}{a}-1}{b-\frac{1}{a}}=\frac{\frac{a}{b}-1}{a-\frac{1}{b}} \implies \frac{b-a}{ab-1}=\frac{a-b}{1-ab} \]which is true.

Yet another complex bash sol, that i somehow quickly despite having zero practice. Let $(ABC)$ be unit circle then $o=0$, $h=\frac{1}{2} \left( a+b+c-\frac{bc}{a} \right)$, $p=\frac{a+\frac{c^2}{a}}{2}$ and $m=\frac{a+b}{2}$, so we need $h,p,m$ colinear but by the composed shifting $X \to (2X-a)a$ and colinearity lemma all we need is $\frac{c^2-ab}{c^2+bc-ab-ac} \in \mathbb R$. But this is easy to prove by taking the conjugates and multiplying numerator and denominator by $abc^2$ thus $HP$ bisects $AB$ as desired, and done .

Direct