Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

Let $O$ be the center of the circumcircle of $A_1A_2\cdots A_{2n}$.

Let $x$ denote the distance from $O$ to any side of $A_1A_2\cdots A_{2n}$ (since $x$ is constant for a regular polygon, there is no ambiguity.)

It is well known that $PH_i=2x$.

Let $M_1,M_2$ be the midpoints of $A_1A_2$ and $A_3A_4$, respectively.

Since $OM_1\parallel PH_1, OM_2\parallel PH_2$. Hence $\angle H_1PH_2 = \angle M_1OM_2=\angle M_1OA_2 + \angle A_2OA_3 +\angle A_3OM_2 = \frac{360^\circ}{n}$.

Similarly, $\angle H_iPH_{i+1}=\frac{360^\circ}{n}$ for any $i$ in modulo $n$. Also $PH_i=PH_{i+1}=2x$ , which implies all $H_iH_{i+1}$ are equal. Hence $H_1H_2\cdots H_n$ is a regular $n$-gon.