Let $N$ be the midpoint of $AB$ note that $NA=NB=NP$ so $\angle PNA = 180^{\circ}-2\angle NAP= 180^{\circ} - A$ so $NP \parallel AC \parallel MN$ so $N,P,M$ are collinear. Now by Ceva's theorem on triangle $AMB$ given that cevians $BQ,MN,AD$ concur we have $$\dfrac{BD}{DM} \cdot \dfrac{MQ}{QA} = 1 \rightarrow \dfrac{BD}{DM} = \dfrac{QA}{MQ}.$$Therefore $DQ \parallel AB.$

Let $X=AD\cap \odot (ABC)$ $\Longrightarrow$ $XB=XC$ $\Longrightarrow$ $XM\perp BM$, so from $\measuredangle XPB=\measuredangle XMB=90^{\circ}$ we get $BPMX$ is cyclic $\Longrightarrow$ $\measuredangle MPD=\measuredangle XBD=\measuredangle DAC$ $\Longrightarrow$ $MP\parallel AC$, since $M$ is midpoint of $BC$ $\Longrightarrow$ $MP$ through by the midpoint of $AB$ which is sufficient to prove that $QD\parallel AB$.

Dear Mathlinkers, Reim's theorem help... Sincerely Jean-Louis

Can you post the Reim's theorem involving solution please.