Let $AD$ cut $\odot(AMN)$ for a second time at $T.$ By symmetry, $AT$ is a diameter of $\odot(AMN) \implies \angle APT = \angle ADB = 90^{\circ}.$ Therefore, $\triangle APT \sim \triangle ADB \implies \tfrac{AP}{AT} = \tfrac{AD}{AB}.$ Similarly, $\tfrac{AQ}{AT} = \tfrac{AD}{AC}$, and it follows that $\tfrac{AP}{AQ} = \tfrac{AC}{AB} \; (\star).$ Meanwhile, Power of a Point yields $BP \cdot BA = BM \cdot BN$ and $CQ \cdot CA = CM \cdot CN.$ As $M$ is the midpoint of $\overline{BC}$, it follows that $\tfrac{BP}{BN} \cdot \tfrac{CN}{CQ} = -\tfrac{AC}{AB}.$ Then according to $(\star)$, we obtain $\tfrac{BP}{BN} \cdot \tfrac{CN}{CQ} \cdot \tfrac{AQ}{AP} = -1$, so by Ceva's Theorem, $AN, BQ, CP$ are concurrent, as desired.

It is also possible to discover $\frac{|AP|}{|AQ|}=\frac{|AC|}{|AB|}$ by considering the power of $A$ with respect to circle $(BPCQ)$, whose existence follows from the fact that $PQ$ and $BC$ are antiparallels induced by $(AMN)$ whose diameter lies on the $A$-altitude by definition.

Remark: Ridiculously similar configuration has appeared in IMO Shortlist 2019 G1. Careful reader will note that $AN$ is tangent to $\odot(BPN)$ and $AM$ is tangent to $\odot(CQM)$, while in shortlist problem intersection of these tangents where defined as point $T$ and it was asked to prove that $AT \parallel BC$. Since $N$ is reflection of $M$ over $D$ we have that $A=T$ and thus we arrive to configuration in this problem. From Ceva's theorem it is sufficient to show that: $$ \frac{AP}{BP} \cdot \frac{BN}{CN} \cdot \frac{CQ}{AQ} = 1\quad (1) $$From PoP with respect to $\odot(APNMQ)$ we have that: $$ BP \cdot BA = BN \cdot BM \quad \text{and} \quad CQ \cdot CA =CN \cdot CM$$This is equivalent to: $$ \frac{ BA}{BM} = \frac{BN}{BP} \quad \text{and} \quad \frac{CQ}{CN} = \frac{CM}{CA} $$Thus (1) becomes: $$ \frac{AP}{AQ} = \frac{AC}{AB} $$The last relation is equivalent to $\triangle APQ \sim \triangle ACB$ or quadrilateral $BPQC$ being cyclic. Observe that $\angle AMN = \angle ANM = \alpha $ and then we have that $\angle ANB = 180^{\circ} - \alpha = \angle APN$. This implies that $\triangle APN \sim \triangle ANB $. Consequently we have that: $$ \angle ANP = \angle ABN = \angle AQP $$Thus quadrilateral $BPQC$ is cyclic as desired.

We have $AN=AM$ as $ND=DM$ and $\measuredangle ADN=90^\circ$. Claim. $BPQC$ is cyclic. Proof. Note that $\measuredangle ANB=\measuredangle ANM=\measuredangle NMA=\measuredangle NPA$, thus $\triangle APN\sim\triangle ANB$. Similarly, $\triangle AMQ\sim\triangle AMC$. Thus, $$AP\cdot AB=AN^2=AM^2=AQ\cdot AC.$$ Let $R=PQ\cap BC$. Hence, $RB\cdot RC=RP\cdot RQ=RN\cdot RM$. Claim. $(R,N,B,C)=-1$. Proof. Note that $(R,N,B,C)=-1$ is equivalent to $RB\cdot NC= RC\cdot BN$. We use $RB\cdot RC=RN\cdot RM$, thus \begin{align*} RB\cdot NC=RB\cdot(RC-RN)=RN\cdot RM-RN\cdot RB=RN\cdot BM\\=RN\cdot MC=RN\cdot (RC-RM)=RN\cdot RC-RN\cdot RM\\=RN\cdot RC-RB\cdot RC=RC\cdot (RN-RB)=RC\cdot BN. \end{align*}From the claim above, we conclude that $AN,BQ,CP$ are concurrent. [asy][asy] size(7cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair O,A,B,C,D,M,N,P,Q,R; O=(0,0);A=dir(110);B=dir(200);C=dir(340);D=foot(A,B,C);M=midpoint(B--C);N=2D-M;path w=circumcircle(A,M,N);P=intersectionpoints(w,A--B)[1];Q=intersectionpoints(w,A--C)[0]; R=extension(P,Q,B,C);path q=circumcircle(B,P,Q); draw(A--B--C--cycle,deep);draw(w,med);draw(A--N,light);draw(B--Q,light);draw(C--P,light); draw(Q--R--B,deep);draw(circumcircle(B,P,Q),light+dashed); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$M$",M,dir(M)); dot("$N$",N,dir(N)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$R$",R,dir(R)); [/asy][/asy]