If you extend $AP$ to meet $w_P$ at $P'$, and let $l_P, l_Q$ be tangents at $A$ in obvious way, $P'X \parallel l_P \implies PM \parallel l_Q$. Then invert about $A$ swapping $B, C$ and the result follows as it preserves point at infinity of corresponding lines, fyi $XY \mapsto MN$

It is well-known that for any pair of isogonal conjugate $P,Q$ with respect to $\triangle ABC$,$\angle BPC+\angle BQC=180^\circ+\angle BAC$,etc. Since $\angle BMC+\angle BNC=\angle BPC+\angle BQC=180^\circ+\angle BAC$ and,without loss of generality, \begin{align*}\angle AMC+\angle ANC& =360^\circ-(90^\circ+\angle BMP+\angle BMC)+(90^\circ+\angle CNQ)\\& =360^\circ-(90^\circ+\angle BCP+\angle BPC)+(90^\circ+\angle CBQ)\\& =360^\circ-(90^\circ+180^\circ-\angle CBP)+(90^\circ+\angle ABP)\\& =180^\circ+\angle CBP+\angle ABP\\& =180^\circ+\angle ABC\end{align*} we obtain that $M,N$ are isogonal conjugate with respect to $\triangle ABC$,which follows that $\triangle APM\sim \triangle AQN$,and then \begin{align*}\frac{PM}{QN}& =\frac{AP}{AQ}\\& =\frac{\sin \angle ABP\cdot BP}{\sin \angle BAP}\cdot \frac{\sin \angle CAQ}{\sin \angle ACQ\cdot CQ}\\& =\frac{\sin \angle CBQ}{CQ}\cdot \frac{BP}{\sin \angle BCP}\\& =\frac{\sin \angle BQC}{BC}\cdot \frac{BC}{\sin \angle BPC}\\& =\frac{\sin \angle BQC}{\sin \angle BPC}\\& =\frac{PX}{QY}\end{align*} so $\triangle XPM\sim \triangle YQN$,and hence $\frac{AM}{AN}=\frac{PM}{QN}=\frac{MX}{NY}$,therefore $MN\parallel XY$,as desired. $Q.E.D.$

It is easy when ones known $HM$ point Clearly $N$ is the the $A-HM$ point of $\triangle ABC$ and through angle chasing one can show $M$ is the midpoint of the symmedian chord of $(ABC)$, $X$ is the intersection of the tangents at $B$ and $C$ and $Y$ is a point such that $ABYC$ is a $\parallel$gm. Consider an Inversion $\phi$ centered at $A$ with radius $\sqrt{AB.AC}$ followed by the reflection along the $A$ angle bisector. $\phi :(ABC) \longrightarrow BC$ and $\phi : (BHC) \longrightarrow (BOC)$ Hence $M \longrightarrow Y$ and $N \longrightarrow X \implies AM.AY=AN.AX \implies \frac{AM}{AX}=\frac{AN}{AY} \implies MN \parallel XY$.

We claim that under a root bc inversion, $M$ swaps with $Y$ and $N$ swaps with $X$. This solves the problem by angle chase under inversion. It is well known that given isogonal conjugates $P, Q$ in $\triangle ABC$, the circumcircles $(BPC)$ and $(BQC)$ swap under root bc inversion. Thus, the image of $M$ under root bc inversion (let this be $M^{*}$) must lie on $(BQC)$. Similarly, $N^{*}$ must lie on $(BPC)$. Extend $AP$ to meet $(BPC)$ again at $Q^{*}$, the image of $Q$ under root bc and extend $AQ$ to meet $(BQC)$ again at $P^{*}$, the image of $P$ under root bc. Angle chasing, we see that $90 = \angle AMP = \angle AP^{*}M^{*} = \angle QP^{*}M^{*}$ and since $M^{*} \in (BQC)$, we conclude that $M^{*}$ is the antipode of $Q$ on $(BQC)$. Similarly, $N^{*}$ is the antipode of $P$ on $(BPC)$. Notice that $X$ and $Y$ are defined as the antipodes of $P$ in $(BPC)$ and $Q$ in $(BQC)$ respectively, so we have proven the claim. Thus, $\angle ANM = \angle AYX$, finishing the problem.

Claim: $M$ and $N$ are isogonal conjugates. Proof. Redefine $N$ as the isogonal conjugate of $M.$ It suffices to prove $N=(BQC)\cap(AQ).$ First, we claim $CBQN$ is cyclic. Indeed, $$\measuredangle BPC+\measuredangle BQC=\measuredangle BAC+180=\measuredangle BMC+\measuredangle BNC$$and $\measuredangle BPC=\measuredangle BMC.$ Then, \begin{align*}&\measuredangle MCP=\measuredangle QCN=\measuredangle QBN\\&\implies \measuredangle CBA-\measuredangle NBA-\measuredangle CBQ+\measuredangle MAC=\measuredangle ACB-\measuredangle PCB-\measuredangle ACM+\measuredangle BAN\\&\implies \measuredangle CBA+\measuredangle MAC+\measuredangle ACM-\measuredangle CBQ=\measuredangle ACB+\measuredangle BAN+\measuredangle NBA-\measuredangle PCB\\&\implies \measuredangle CBA+180-\measuredangle CMA-\measuredangle CNQ=\measuredangle ACB+180-\measuredangle ANB-\measuredangle PMB\\&\implies \measuredangle QNA=\measuredangle AMP=90.\end{align*}$\blacksquare$ Take the $\sqrt{bc}$ inversion, denoting $Z^*$ as the inverse of $Z.$ Claim: $\omega_P^*=\omega_Q.$ Proof. Let $P'=\overline{AQ}\cap\omega_Q$ and note $$\measuredangle QP'B=\measuredangle QCB=\measuredangle ACP$$so $\triangle AP'B\sim\triangle ACP$ or $P^*=P'.$ $\blacksquare$ Hence, $P$ and $Q$ lie on $\omega_Q^*$ and $\omega_P^*,$ respectively. Also, $\measuredangle N^*Q^*A=\measuredangle QNA=90$ so $\overline{PN^*}$ is the diameter of $\omega_Q^*.$ Hence, $Y^*$ is the foot from $P$ to $\overline{AN^*}=\overline{AM}$ so $Y^*=M.$ Similarly, $X^*=N.$ Thus, $$AX^*\cdot AN^*=AB\cdot AC=AY^*\cdot AM^*$$and $\triangle AX^*Y^*\sim\triangle AM^*N^*.$ $\square$