1. Identify Key Points and Properties:
- Given triangle $\triangle ABC$ with $AB < AC$.
- $D$ is the point where the angle bisector of $\angle BAC$ meets $BC$.
- $M$ is the midpoint of $BC$.
- $P$ is the foot of the perpendicular from $B$ to $AD$.
- $Q$ is the intersection of $BP$ and $AM$.

2. **Introduce Point $X$:**
- Let $X$ be the point where $AD$ intersects the circumcircle $\odot (ABC)$.
- Since $X$ lies on the circumcircle, $XB = XC$ (by the property of the circumcircle).

3. Perpendicularity and Cyclic Quadrilateral:
- Since $M$ is the midpoint of $BC$, $XM$ is perpendicular to $BC$ (as $X$ is on the circumcircle and $M$ is the midpoint).
- Therefore, $\angle XMB = 90^\circ$.
- Given $P$ is the foot of the perpendicular from $B$ to $AD$, $\angle XPB = 90^\circ$.
- Hence, quadrilateral $BPMX$ is cyclic (as opposite angles sum to $180^\circ$).

4. Parallelism and Angle Chasing:
- In cyclic quadrilateral $BPMX$, $\angle MPD = \angle XBD$.
- Since $X$ lies on the circumcircle, $\angle XBD = \angle DAC$ (by the Inscribed Angle Theorem).
- Therefore, $\angle MPD = \angle DAC$.

5. Conclusion of Parallelism:
- Since $\angle MPD = \angle DAC$, it follows that $MP \parallel AC$.
- Given $M$ is the midpoint of $BC$, line $MP$ passes through the midpoint of $AB$.
- This implies that $QD \parallel AB$.

The final answer is $\boxed{ DQ \parallel AB }$.