Let $O$ be the center of the circumscribed circle around $\triangle ABC$. We will show that each of the lines $p_A, p_B, p_C$ passes through $O$.

Because of symmetry, it is enough to show that $OC \perp H_A H_B$.

Let $D$ be the point of intersection of these two lines. We restrict ourselves to the case where $\triangle ABC$ is acute (since in the case of $\triangle ABC$ being obtuse the argument is analogous). It is enough to use the fact that $\angle H_A CD = \angle BCO = 90^\circ - \alpha$ and $\angle DH_A C = \angle H_B H_A C = \alpha$ (the last equality follows from the fact that $ABH_A H_B$ is inscribed).

According to Carnot's theorem, it is sufficient (and necessary) to show that
$$|AH_B|^2 - |AH_C|^2 + |BH_C|^2 - |BH_A|^2 + |CH_A|^2 - |CH_B|^2 = 0.$$
For that purpose, it is sufficient to sum the obvious equalities: $|BH_C|^2 - |AH_C|^2 = |BC|^2 - |AC|^2$, $|CH_A|^2 - |BH_A|^2 = |AC|^2 - |AB|^2$, and $|AH_B|^2 - |CH_B|^2 = |AB|^2 - |BC|^2$.