Let $O$ denote the center of the circumcircle $\omega$ and denote $\alpha := \angle BAC$, $\beta := \angle ABC$ and $\gamma := \angle BCA$. Let $A' := (AO) \cap \omega$. First we show that $N \in A'H$.

Indeed, let $N' := AH \cap BC$. Since $O$ is the midpoint of $AA'$, the centroid $G$ of the triangle $AHA'$ is the point on $OH$ satisfying $HG = 2GO$. By Euler's theorem, $G$ is also the centroid of the triangle $ABC$. It follows that $HBA'C$ is a parallelogram and thus $N' = N$. Now using the fact that $PEHK$, $BDHE$ and $PKCA$ are circumscribed, we compute

$$\begin{align*} \angle PED &= \angle PEH - \angle DEH \\ &= (180^\circ - \angle PKH) - \angle DBH \\ &= (180^\circ - (\angle PKC - 90^\circ)) - (90^\circ - \gamma) \\ &= 180^\circ - \angle PKC + \gamma \\ &= \angle PAC + \gamma \\ &= \angle PAB + \alpha + \gamma. \end{align*}$$

Similarly, using the fact that $PQAL$, $QDHL$ and $LEHA$ are circumscribed, we compute

$$\begin{align*} \angle PQD &= \angle PQL + \angle LQD \\ &= \angle PAL + (180^\circ - \angle LHD) \\ &= \angle PAL + (180^\circ - (\angle LHE + \angle EHD)) \\ &= \angle PAL - \angle LHE + \beta \\ &= \angle PAL - \angle LAE + \beta \\ &= \beta - \angle PAB. \end{align*}$$

It follows that $\angle PED + \angle PQD = \alpha + \beta + \gamma = 180^\circ$, hence $DEPQ$ is circumscribed.