Let $H$ be the orthocenter of the triangle $ABC$. Quadrilaterals $BIHK$ and $CIHJ$ are cyclic, so $\angle HIK = \angle HBK = \angle ABJ = \angle ACK = \angle JCH = \angle HIJ$.

Therefore, $IA$ is the angle bisector of $JIK$. Since $IA \perp BC$ and $MP \parallel BC$, it follows that $IA \perp MP$. Because $IA$ is both an angle bisector and an altitude in $\triangle IMP$, this triangle is isosceles and $A$ is the midpoint of $MP$. (*)

Let $S$ be the reflection point of $A$ with respect to point $D$. Since $D$ is the midpoint of $BC$, it follows that $ABSC$ is a parallelogram. Consequently, $BS \parallel AC$ and, since $BJ \perp AC$, we deduce that $BJ \perp BS$. From $\angle SBN = \angle SAN = 90^\circ$ it follows that $SBAN$ is cyclic, therefore $\angle NSA = \angle ABN = 90^\circ - \angle A$. Similarly, the quadrilateral $SCAQ$ is cyclic, hence $\angle QSA = \angle ACQ = 90^\circ - \angle A = \angle NSA$. In the triangle $SNQ$, the altitude $SA$ is also an angle bisector, therefore $A$ is the midpoint of $NQ$. Since the diagonals of the quadrilateral $MNPQ$ bisect each other, it follows that $MNPQ$ is a parallelogram.

Alternative solution for ()*. Let $E$ be the projection of $D$ onto $AC$.
Since $\angle DEA = \angle AJN = 90^\circ$ and $\angle DAE = 90^\circ - \angle NAJ = \angle ANJ$, it follows that $\triangle DAE \sim \triangle ANJ$. Thus, $\frac{NA}{AD} = \frac{AJ}{DE} = 2 \cdot \frac{AJ}{BJ}$. Similarly, we obtain $\frac{QA}{AD} = 2 \cdot \frac{AK}{CK}$. The triangles $ABJ$ and $ACK$ are similar (AA), thus $\frac{AJ}{BJ} = \frac{AK}{CK}$, and consequently $AQ = AN$.