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Triangles $AEB$ and $BED$ are similar since $\angle BAE=\angle EAC=\angle DBE$. Hence $\angle AEK=\angle BEL$ as the angles between a median and a side in similar triangles. Denote these angles by $\phi$. Then $\angle EKL=\phi$ since $KL$ is a midline of $△ABD$.

Analogously, let $\psi =\angle AEN=\angle CEM=\angle ENM$. And let $\beta =\angle ABC$, $\gamma =\angle ACB$.

The triangle $PEL$ is isosceles, therefore $\angle PEL=90^{\circ }-\frac{1}{2}\angle EPL=90^{\circ }-\angle EKL=90^{\circ }-\phi$ and

$$\angle PEA=\angle PEL-\angle AEL=\angle PEL-(\angle AEB-\angle BEL)=90^{\circ }-\phi -(\gamma -\phi )=90^{\circ }-\gamma .$$

Analogously $\angle QEA=90^{\circ }-\beta$.

Thus $\angle PEQ=\angle PEA+\angle QEA=180^{\circ }-\beta -\gamma =\angle BAC$.