Since $Q$ is the Miquel point of $\triangle ABC \cup MN,$ it follows that $P,Q,M,B$ are concyclic. Let $R \equiv PQ \cap AC$ and assume that $\overline{PQR}$ is tangent to the circumcircle $(O).$ Then $\angle RQA=\angle QBM=\angle QPM$ $\Longrightarrow$ $QA \parallel PN.$ Hence, the cyclic quadrilateral $AQMN$ is an isosceles trapezoid with bases $AQ$ and $MN$ $\Longrightarrow$ perpendicular bisector of $\overline{AQ}$ is also the perpendicular bisector of $\overline{MN}$ $\Longrightarrow$ $OM=ON.$ The converse can be proved with exactly the same arguments.

This problem is nice.Here's a pure synthetic solution. 1.$PQ$ tangent to $\omega \Rightarrow OM=ON$. Since $PQ$ is tangent to $\omega$ we have $\angle{CQP}=\angle{QBC}=\angle{QBP}$.Also note that $\angle{CQP}=\angle{QAC}=\angle{QAN}=\angle{QMN}=\angle{QMP}$.Thus $\angle{QMP}=\angle{QBP} \implies QMBP$ is concyclic.So $\angle{NCQ}=\angle{ACQ}=\angle{ABQ}=\angle{QBM}=\angle{QPM}=\angle{QPN} \implies QNCP$ is cyclic.Hence $\angle{CQP}=\angle{CNP}=\angle{ANM}=\angle{QAN}$.So $AQ \parallel MN \implies AQNM$ is an isosceles trapezoid.Thus $QM=AN$ and $AM=QN$. Now note that $\angle{AMQ}=\angle{ANQ} \Rightarrow \angle{BMQ}=\angle{CNQ}$.Also $\angle{MBQ}=\angle{ABQ}=\angle{ACQ}=\angle{NCQ}$.Thus $\triangle{BMQ} \sim \triangle{CNQ} \implies \frac{BM}{CN}=\frac{MQ}{NQ} \implies BM \times NQ=CN \times QM \implies BM \times AM=CN \times AN$. So since power of $M$ and $N$ are equal wrt $\omega$ we have $OM=ON$. 2.$OM=ON \Rightarrow PQ$ is tangent to $\omega$. As with the above situation it follows that $\triangle{BMQ} \sim \triangle{CNQ} \Rightarrow \frac{BM}{CN}=\frac{MQ}{NQ} \Rightarrow BM \times NQ=CN \times MQ$. Also by hypothesis we have $BM \times AM=CN \times AN$.Combining these yeilds $AN \times NQ=AM \times MQ \Rightarrow [AMQ]=[ANQ] \Rightarrow [AMZ]=[QNZ] \Rightarrow AZ \times MZ=QZ \times NZ$. We also have $\frac{AZ}{MZ}=\frac{QZ}{NZ}$.Summing these it follows that $AN=QM$ or $AQNM$ is an isosceles trapezoid. Also note that as with the previous situation we similarly have $QNCP$ cyclic.Thus $\angle{PQC}=\angle{PNC}=\angle{ANM}=\angle{NAQ}=\angle{CAQ}$ or $PQ$ is tangent to $\omega$ as required.

[asy][asy] unitsize(100); pen n_purple = rgb(0.7,0.4,1), n_blue = rgb(0,0.6,1), n_green = rgb(0,0.4,0), n_orange = rgb(1,0.4,0.1), n_red = rgb(1,0.2,0.4); pair A, B, C, O, M, N, Q, P; A = dir(110); B = dir(210); C = dir(330); O = origin; M = intersectionpoints(circle(O, 0.7), A--B)[0]; N = intersectionpoints(circle(O, 0.7), A--C)[1]; Q = reflect(O, circumcenter(A, M, N)) * A; P = extension(M, N, B, C); draw(P--Q, gray(0.6)); draw(C--P, gray(0.6)); draw(M--P, gray(0.6)); draw(A--Q, gray(0.6)); draw(A--B--C--cycle); draw(circumcircle(A, M, N), n_red); draw(unitcircle, n_blue); dot(A^^B^^C^^M^^N^^P^^Q); label("$A$", A, dir(130)); label("$B$", B, dir(210)); label("$C$", C, dir(290)); label("$M$", M, dir(180)); label("$N$", N, dir(70)); label("$Q$", Q, dir(340)); label("$P$", P, dir(310)); [/asy][/asy] By angle chasing, $\overline{PQ}$ is tangent to $\omega$ iff $\overline{AQ} \parallel \overline{MN}$, so it suffices to show $OM = ON$ iff $\overline{AQ} \parallel \overline{MN}$. Lemma. Given two circles $\omega_1$ and $\omega_2$, the locus of points $P$ such that $\text{Pow}_{\omega_1}(P) - \text{Pow}_{\omega_2}(P)$ is constant is a line perpendicular to the line joining the centers of $\omega_1$ and $\omega_2$. Proof. $\text{Pow}_{\omega_1}(P) - \text{Pow}_{\omega_2}(P)$ is linear in $P$. Now $\overline{MN}$ is parallel to $\overline{AQ}$, the radical axis of $\omega$ and $(AMN)$, iff $\text{Pow}_{\omega}(M) - \text{Pow}_{(AMN)}(M) = \text{Pow}_{\omega}(N) - \text{Pow}_{(AMN)}(N)$ iff $\text{Pow}_{\omega}(M) = \text{Pow}_{\omega}(N)$ iff $OM = ON$.

Notice that $\frac{PB}{PC} = \left(\frac{QB}{QC}\right)^2 = \left(\frac{BM}{CN}\right)^2$. By Menelaus' theorem on $\triangle ABC$ and $\overline{PMN}$, we have $\frac{PB}{PC} = \frac{AN}{AM} \cdot \frac{BM}{CN}$. Combining the two equations yields $\frac{AN}{BM} = \frac{AM}{CN}$, which is equivalent to $BM \cdot AM = CN \cdot AN \implies \text{Pow}(M, \odot(O)) = \text{Pow}(N, \odot(O))$ so $OM = ON$.

Let $\gamma$ denote $(AQMN)$. First, assume $PQ$ is tangent to $\omega$. Since $Q$ is the Miquel Point of complete quadrilateral $BCNM$, we know $BMQP$ is cyclic. Thus, we have $$\measuredangle NMA = \measuredangle PMB = \measuredangle PQB = \measuredangle QAB = \measuredangle QAM$$so $AQ \parallel MN$. Hence, $AQMN$ is a (cyclic) isosceles trapezoid. Properties of Miquel Points yield $QBM \overset{+}{\sim} QCN$, so $$\frac{AM}{AN} = \frac{QN}{QM} = \frac{CN}{BM}$$which gives $$Pow_{\omega}(M) = AM \cdot BM = AN \cdot CN = Pow_{\omega}(N)$$clearly implying $OM = ON$. Now, assume $OM = ON$. Reversing our logic clearly gives $$\frac{AM}{AN} = \frac{CN}{BM} = \frac{QN}{QM}.$$Since $M$ is in between $A$ and $B$, and $N$ is in between $A$ and $C$, we know that neither of these points can lie outside $(ABC)$. Because one of the arcs of $\gamma$ bounded by $A$ and $Q$ must travel outside $(ABC)$, it follows that $A$ and $Q$ are adjacent vertices of quadrilateral $AQMN$. Since there exists exactly one point $Q$ on $\gamma$ which is on the same side of $MN$ as $A$ such that $$\frac{AM}{AN} = \frac{QN}{QM}$$it follows that $Q$ is the unique point on $\gamma$ such that $AQ \parallel MN$. Now, angle chasing clearly implies the desired tangency, so we're done. $\blacksquare$ Remarks: My solution is unnecessarily long and not generalizable because of configuration issues. Recognizing the perpendicular bisector argument is much cleaner. Overall, this was a pretty easy problem.

Very nice problem. The condition is equivalent to $\frac{BM}{CN} = \frac{AN}{AM}$ by Power of a Point, which rewrites as $\triangle QMN \sim \triangle ANM$ by considering $Q$ as the Miquel point. On the other hand, this implies $AQNM$ is an isosceles trapezoid. As $QNCP$ is also cyclic, we have $$\measuredangle PQC = \measuredangle QNM = \measuredangle QAC,$$which is equivalent to the tangency.