We will use the following well-known fact: For a $\triangle ABC$ and a point $D$ on $AB$, $CD$ is tangent to $(ABC)$ if and only if $\frac{DA}{DB}=\frac{CA^2}{CB^2}$. We claim that $CM$ and $CN$ are isogonal with respect to $\angle ACB$. By Menelaus theorem, $\frac{DA}{DB}\frac{KB}{KC}\frac{LC}{LA}=1$, so Thales yields$\frac{KB}{KC}\frac{LC}{LA}=\frac{BM}{MA}\frac{AN}{NB}=\frac{CB^2}{CA^2}$. Let $\angle BCN=\varphi_1, \angle ACM=\varphi_2$. After applying Law of Sines in $\triangle ACN$, $\triangle BCM$, $\triangle ACM$, $\triangle BCN$ combined with the above equation yields that $\sin (\gamma-\varphi_1)\sin(\varphi_2)=\sin(\gamma-\varphi_2)\sin(\varphi_1)$. Using the formula $2\sin \alpha \sin \beta =\cos(\alpha-\beta)-\cos(\alpha+\beta)$ twice, we get that $\cos(\gamma-\varphi_1+\varphi_2)=\cos(\gamma-\varphi_2+\varphi_1)$, so $\varphi_1=\varphi_2=\varphi$. Notice that $\angle QMN=\angle QPN=\angle QCB$, so $QMCB$ is cyclic; similarly, $AQNC$ is cyclic. Since $\angle MQN=180^{o}-\gamma$, we easily obtain that $\angle BQN =\angle AQM =\varphi$. Thus $(MQN)$ and $(AQB)$ touch at $Q$, so we want $DQ$ to touch $(AQB)$. Using the fact in the beginning of the solution, we want $\frac{CA^2}{CB^2}=\frac{DA}{DB}=\frac{QA^2}{QB^2} \iff \frac{QA}{QB}=\frac{CA}{CB}$. By law of sines in $\triangle AQB$, $\frac{QA}{QB}=\frac{\sin \angle ABQ}{\sin \angle BAQ}=\frac{\sin \angle MCQ}{\sin \angle NCQ}$. Let $CP \cap AB=R, CP \cap KL=S$. By ratio lemma in $\triangle CMN$, $\frac{RM}{RN}\frac{CN}{CM}=\frac{\sin \angle MCQ}{\sin \angle NCQ}$, so we want $\frac{CA}{CB}=\frac{RM}{RN}\frac{CN}{CM}$. By power of point, $RM.RB=RQ.RC=RN.RA \iff \frac{RM}{RN}=\frac{RA}{RB}$. Using ratio lemma in $\triangle ABC$, $\frac{RM}{RN}=\frac{RA}{RB}=\frac{CA}{CB}\frac{\sin \angle ACR}{\sin \angle BCR}$, so we want $\frac{CM}{CN}=\frac{\sin \angle ACR}{\sin \angle BCR}$. By ratio lemma in $\triangle CKL$ using the fact that $S$ is the midpoint of $KL$ (because $CKPL$ is a parallelogram), we get that $\frac{\sin \angle ACR}{\sin \angle BCR}=\frac{CK}{CL}$, so we want $\frac{CM}{CN}=\frac{CK}{CL}$. Now, this easily follows from the similarity of $\triangle CMK$ and $\triangle CNL$ (which is true since $\angle CKM=\angle CLN$, $\angle MCK=\angle NCL$).

Here is my solution from contest that I came up with because I forgot how to deal with parallel lines

Note that $DM = \frac{DB \cdot DK}{DL}$ and $DN = \frac{DA \cdot DL}{DK}$ so $DM \cdot DN = DB \cdot DA = DC^2$. So $D$ lies on the radical axis of $(AQB)$ and $(MNPQ)$, meaning it suffices to show that these two circles are tangent. We have $\angle A + \angle BCM = \angle DCM = \angle CNM = \angle A + \angle ACN$ so $\angle BCM = \angle ACN$, so $CN$ and $CM$ are isogonal in $\angle ACB$. Finally, observe that $\angle BNQ = \angle MNQ = \angle MPQ = \angle BCQ$, giving that points $B,C,N,Q$ are cyclic. Analogously, points $A,C,M,Q$ are concyclic as well. For the tangency, it is enough to show that $\angle BQM + \angle BAQ = \angle BNQ$. But $\angle BQM = \angle BQN - \angle MQN = 180 - \angle BCN - (180 - \angle C) = \angle ACN$, and $\angle BAQ = \angle MAQ = \angle MCQ$. So $\angle BQM + \angle BAQ = \angle ACN + \angle MCQ = \angle BCM + \angle MCQ = \angle BCQ = \angle BNQ$, as desired, so we're done. $\blacksquare$

Note that $DN \cdot DM = DA \cdot DB$, so $DC$ is tangent to $(CMN)$, and $CN, CM$ are isogonal wrt $\angle{C}$. By PoP at $D$, it suffices to show that $\frac{QN}{QM} = \frac{CN}{CM}$. If $Q' = CP \cap (ABC)$ then $\triangle{PNM} \sim \triangle{CAB}$ so a homothety at $CP \cap AB$ takes $N, M, P, Q$ to $A, B, C, Q'$, thus $\frac{QN}{QM} = \frac{Q'A}{Q'B}$, which by the ratio lemma equals $\frac{AQ}{BQ} \div \frac{AC}{BC}$. $\frac{AQ}{BQ} = \frac{AN}{BM}$ by homothety, so this is equal to $\frac{AN}{BM} \div \frac{AC}{BC}$ which is also equal to $\frac{CN}{CM}$ by LoS on $\triangle{CAN}, \triangle{CBM}$. $\square$

Since we have $KA \parallel LN$ and $KM\parallel LB$, we obtain that \[ \frac{DA}{DN}=\frac{DK}{DL}=\frac{DM}{DB} \quad \Rightarrow \quad DA\cdot DB = DN\cdot DM. \]So it suffices to show that $DC=DQ$. Let the lines $NL$ and $MK$ meet with the line $DC$ at $Z(.)$ and $Y(.)$, respectively. Some angle chasing work gives us that $\angle NZC=\angle NBC=\angle NMY$ and $\angle MYC = \angle MAC =\angle MNZ$, and so the quadrilaterals $AYCM$, $BZCN$ and $NYZM$ are cyclic. Moreover, $\angle NQC = \angle NQP=\angle CMP = \angle NMY$ and $Q$ lies on the circle $(BZCN)$. Similarly, we can show that $Q$ also lies on the circle $(AYCM)$. Now, considering an inversion with a centre $D$ and radius $CQ$, we can see that the circle $(AYCM)$ moves to the circle $(BZCN)$ and vice versa. So $Q$ should be a fixed point, i.e. $DQ^2=DC^2$. We are done!

Guess and prove that $DA \cdot DB = DM \cdot DN$ (make sense that there are a lot of common tangents and powers ) by Menelaus and Thales and then trig bash to show $CQ = 2CD\cos\angle DCQ$, which implies $DQ = DC$ and we are done.

Similar triangles give $$DM\cdot DN = \left(DB\cdot \frac{DK}{DL}\right)\left(DA\cdot \frac{DL}{DK}\right) = DB\cdot DA = DC^2$$ so $DC$ is tangent to $(CMN)$. In addition, angle chasing gives $$\angle MQC = \angle MQP = \angle MNP = \angle MAC$$ so $ACMQ$ is cyclic, and similarly $BCNQ$ is cyclic. Now invert the diagram about $C$, letting $\text{arg}^*$ denote the image of a point after the inversion. We see that $M^*N^*\parallel A^*B^* \parallel C^*D^*$ are parallel chords in $(A^*B^*C^*)$ and $Q^* = (A^*M^*\cap B^*N^*)$. But then $A^*B^*D^*C^*$ and $A^*B^*M^*N^*$ are $\text{CyclicISLscelesTrapezoid}^{\text{TM}}\text{s}$, so $Q^*C^* = Q^*D^*$. This translates to $DQ = DC$ in the original diagram, so $$DQ^2 = DC^2 = DM\cdot DN$$ as desired.

There is my solution to the contest - a quite straightforward trig bash! We'll use standard notations for the angles of $\triangle ABC$. WLOG let $BC>AC$. Let the tangent at $Q$ w.r.t. $\omega$ meet $AB$ in $D'$ - we're going to prove that $D \equiv D'$. We have that $(M,N,Q,P) \overset{Q}{=}(M,N,D',S)$, where $S=CQ \cap AB$. Also, let $CS \cap \Omega = Q' \neq C$. Then from $NP \| AC$ and $MR \| BC$ we get that $MPNQ \sim BCAQ' \implies (M,N,Q,P) = (B,A,Q',C)$. We are left to prove that $(B,A,Q',C)=(M,N,D,S)$. Denote $\angle CDL = \phi$. Because of the parallelogram $KPLC, U=KL \cap CP$ is the midpoint of $KL$. Further we'll use the sine law a couple of times. $$(B,A,Q',C)=\frac{BQ'.AC}{BC.AQ'}=\frac{BQ'}{AQ'}.\frac{AC}{BC}=\frac{\sin \angle BCQ'}{\sin \angle ACQ'}.\frac{\sin \beta}{\sin \alpha}=\frac{\sin \angle CLU. \sin \beta}{\sin \angle KCU. \sin \alpha}=\frac{\sin (\alpha - \phi).\sin \beta}{\sin (\beta + \phi) .\sin \alpha}.$$From ratio lemma and law of sines: \[\frac{NS}{MS} = \frac{\sin \angle NPS. \sin \angle PMS}{\sin \angle PSM .\sin \angle PNS}= \frac{\sin \angle ACQ' . \sin \beta}{\sin \angle BCQ' . \sin \alpha} = \frac{\sin (\beta + \phi).\sin \beta}{\sin (\alpha - \phi) . \sin \alpha}\]\[\frac{DM}{AD}=\frac{DM}{DK}.\frac{DK}{AD}=\frac{\sin (\alpha - \phi) . \sin \alpha}{\sin (\beta + \phi).\sin \beta}\]\[\frac{DN}{AD}=\frac{DN}{DL}.\frac{DL}{DB}.\frac{DB}{DC}.\frac{DC}{AD}=\frac{\sin(\beta + \phi)}{\sin\alpha}\frac{\sin \beta}{\sin (\alpha - \phi)}.\frac{\sin \alpha}{\sin \beta}.\frac{\sin \alpha}{\sin\beta}=\frac{\sin(\beta+\phi). \sin \alpha}{\sin(\alpha - \phi). \sin\beta}\] From the last two we obtain that $\frac{DM}{DN}=\frac{\sin ^2(\alpha - \phi)}{\sin ^2(\beta + \phi)}$, so we are ready to evaluate $(M,N,D,S)$. \[(M,N,D,S) = \frac{MD}{ND}.\frac{NS}{MS}=\frac{\sin ^2(\alpha - \phi)}{\sin ^2(\beta + \phi)}.\frac{\sin(\beta + \phi).\sin\beta}{\sin(\alpha - \phi).\sin\alpha}=(B,A,Q',C)\]The last one finishes the proof.

My solution from the contest (ignore that I forgot to post it for over a month): WLOG assume that $\angle A<\angle B$ so $DA>DB$. Let $X = \overline{DC}\cap\overline{NL}$ and $Y = \overline{DC}\cap\overline{MK}$. Quick angle chasing implies that $MQAYC$ and $NQBXC$ are cyclic pentagons and let $O_{1}$ and $O_{2}$ be the centers of their respective circumcircles. Notice that $\triangle AKM \sim \triangle ACB \sim\triangle NLB \sim \triangle YKC \sim \triangle CLX$ due to equal angles. Now we have that $\triangle DLB \sim \triangle DKM\Longrightarrow \frac{KM}{LB} = \frac{DK}{DL}$ and $\triangle DLC \sim \triangle DKY\Longrightarrow \frac{KY}{CL}=\frac{DK}{DL}$, so \[\frac{KM}{LB} = \frac{DK}{DL} = \frac{KY}{CL} \Longrightarrow \frac{KM}{KY} = \frac{BL}{CL}\]The last ratio equality implies that $AMCY\sim NBXC$ as the similarity ratio between $\triangle AKM$ and $\triangle YKC$ $\left(\frac{KM}{KY}\right)$ is equal to the similarity ratio between $\triangle NLB$ and $\triangle CLX$ (which is $\frac{BL}{CL}$). Also from $\triangle DLB\sim\triangle DKM$ and $\triangle DLC\sim\triangle DKY$ we have that $\frac{DX}{DC} = \frac{DB}{DM} = r$. Consider the homothety $h$ centered at $D$ and ratio $r$. Then $h(C)=X$, $h(M)=B$. Also $AK\parallel NL$ so $\triangle DAK\sim\triangle DNL$ hence $\frac{DA}{DN}=\frac{DK}{DL}=\frac{DM}{DB}$, so $h(A)=N$ and similarly $h(Y)=C$ which implies that $h(AMCY)=NBXC$ and furthermore $h(O_{1})=O_{2}$. The last gives that $D,O_{1},O_{2}$ are collinear, so $DC=DQ$ as $\overline{CQ}$ is the radical axis of the two cyclic pentagons and so $\overline{O_{1}O_{2}}$ is the perpendicular bisector of $CQ$. Now recall that $\frac{DA}{DN}=\frac{DK}{DL}=\frac{DM}{DB}$ because of the similarities we mentioned, so \[DQ^2 = DC^2 = DA\cdot DB = DM\cdot DN\Longrightarrow DQ\text{ is tangent to } \omega\]

A pure angle chase solution(sketch).

Let $E$ be the reflection of $C$ in $DK$ Claim 1: $E$ is the miquel point of $LKNM$

Claim 2 $DE$ is tangent to $(EPNM)$ CLaim 3: $D$ is the circumcenter of $CEQ$

Claim 2 finishes.

It's well-known that $\frac{DA}{DB}=\frac{AC^2}{BC^2}$ (can prove this with projective and symmedian+isogonality stuff, probably a better way exists). Then by Menelaus we have $\frac{DA}{DB}\frac{LB}{LC}\frac{KC}{KA}=1$. Using our parallelisms this becomes $\frac{AN\cdot AM}{BN\cdot BM}=\frac{AC^2}{BC^2}$. Note that $AN\cdot AM=\mathrm{Pow}_{(MNP)}(A)-\mathrm{Pow}_{(ABC)}(A)$ and something similar holds for $B$, so by linearity of power we find that $D$ has equal power to $(ABC)$ and $(MNP)$. On the other hand, let $\ell$ be the tangent to $(MNP)$ at $Q$ and let $\ell \cap \overline{CD}=D'$. Homothety and angle chasing imply that $\triangle D'CQ$ is $D'$-isosceles, so $D'$ also lies on the radical axis of $(ABC)$ and $(MNP)$. Thus if $D' \neq D$ the radical axis of $(ABC)$ and $(MNP)$ is $\overline{CD}$, but this is absurd as $C$ doesn't lie on $(MNP)$.

Solved with MathLuis. Our god carried me as always. We let $E=\overline{KM} \cap \overline{CD}$ and $F= \overline{LN} \cap \overline{CD}$. So, we first make some observations in this setup. Claim : Pentagons $CEAQM$ and $CNQBF$ are cyclic. Proof : First note that, \[\measuredangle QCA = \measuredangle QPN = \measuredangle QMN = \measuredangle QMA\]So, in fact $Q$ lies on $(AMC)$. Also, \[\measuredangle CEM = \measuredangle DCB = \measuredangle CAB = \measuredangle CAM\]Thus, $E$ also lies on $(AMC)$. Thus, $CEAQM$ is indeed cyclic as claimed. Similarly, we can also obtain that $CNQBF$ is also cyclic. Claim : $AE \parallel CN$ and $CM \parallel BF$. Proof : Via the previous cyclic quadrilaterals, we have that \[DN \cdot DB = DC \cdot DF \text{ and } DM \cdot DA = DC \cdot DE\]Dividing these two relations gives us that, \begin{align*} \frac{DA}{DN} \cdot \frac{DM}{DB} &= \frac{DE}{DF}\\ \frac{DA}{DN} \cdot \frac{DC}{DF} &= \frac{DE}{DF} \ \ \text{ since $ME \parallel BC$}\\ \frac{DA}{DN} &= \frac{DE}{DC} \end{align*}from which it follows that $AE \parallel CN$. Similarly, we can show that $CM \parallel BF$. Claim : $D$ lies on the radical axis of circles $(NQM)$ and $(CAB)$ Proof : First, notice that by the previously established parallel lines and cyclic quadrilaterals, \[\measuredangle DAE = \measuredangle DNC = \measuredangle BFC = \measuredangle BFE \]Thus, points $A$,$B$,$F$ and $E$ are concyclic. Further, \[\measuredangle NME = \measuredangle NBC = \measuredangle NFC = \measuredangle NFE \]which implies that points $E$,$N$,$M$ and $F$ are concyclic. Now, using these facts we can see that \[DA \cdot DB = DE \cdot DF = DN \cdot DM\]which proves the desired claim. Now, we have one last claim. Claim : Circles $(AQB)$ and $(NQM)$ are tangent. Proof : Note that, \[\measuredangle MQB = \measuredangle NQB + \measuredangle MQN = \measuredangle NCB + \measuredangle BCA = \measuredangle NCA\]Similarly, we can show that, $\measuredangle AQN = \measuredangle BCM$. Then, \[\measuredangle MQB = \measuredangle NCA = \measuredangle CNF = \measuredangle CBF = \measuredangle BCM = \measuredangle MQB\]which implies that indeed circles $(AQB)$ and $(NQM)$ are tangent as desired. Now, this means that the radical axis of $(AQB)$ and $(NQM)$ must be their common tangent. But, we already proved that $D$ lies on this radical axis. Further, $Q$ clearly lies on the radical axis as it is an intersection point of the two circles. Thus, $\overline{DQ}$ is in fact the common tangent to the circles $(AQB)$ and $(MQN)$ from which we can conclude that indeed $DQ$ is tangent to $\omega$, which was the desired conclusion.

First we show $\text{Pow}_{\omega}(D)=\text{Pow}_{\Omega}(D)\iff DN\cdot DM= \frac{DL\cdot DA}{DK}\cdot \frac{DK\cdot DB}{DL}=DA\cdot DB=DC^2.$ As $D$ lies on the radax of $\Omega$ and $\omega$ so we just need to to show $DQ=DC$. First we prove $\odot(MQAC)$, this is true because $\angle MQC=\angle MNP=\angle MAC$, similarly $\odot(NQBC)$. Now a homothety centered at $D$ takes $\odot(MQAC)\to \odot(NQBC)$ because $MK \cap CD\to C, A\to N$ and $M\to B$, letting the center of these circles be $O_1, O_2$ respectively we see that $\overline{D-O_1-O_2}$, also $O_1O_2$ is perpendicular bisector of $CQ$ which further implies $DQ=DC$.

Let $CP\cap AB=T,CP\cap KL=R$. We will prove that $\frac{DN}{DM}\overset{?}{=}\frac{QN^2}{QM^2}$. Note that since $CLPK$ is a parallelogram, $RK=RL$. \[\measuredangle QMN=\measuredangle MPN=\measuredangle QCA \ \ \text{and} \ \ \measuredangle MNQ=\measuredangle MPQ=\measuredangle BCQ\]\[\frac{QM^2}{QN^2}\overset{?}{=}\frac{DM}{DN}=\frac{KM}{LN}.\frac{LP}{KP}\]\[(\frac{KM}{KP}.\frac{CP}{CT}.\frac{AT}{AM})(\frac{LP}{LN}.\frac{BN}{BT}.\frac{CT}{CP})=1\implies \frac{QM^2}{QN^2}\overset{?}{=}\frac{KM}{LN}.\frac{LP}{KP}=\frac{TB}{TA}.\frac{AM}{BN}\]\[\frac{TN}{TM}.\frac{QM^2}{QN^2}\overset{?}{=}\frac{AM}{BN}=\frac{MN.\frac{AK}{NP}}{MN.\frac{LB}{MP}}=\frac{AK}{NP}.\frac{MP}{BL}=\frac{AK}{BL}.\frac{CB}{CA}\]\[\frac{CA}{CB}.\frac{QM}{QN}=\frac{TN}{TM}.\frac{QM^2}{QN^2}\overset{?}{=}\frac{AK}{BL}.\frac{CB}{CA}\implies \frac{AK}{BL}\overset{?}{=}\frac{QM}{QN}.\frac{CA^2}{CB^2}\]We have $\frac{QM}{QN}=\frac{\sin LCR}{\sin RCK}=\frac{CK}{CL}$ since $RL=RK$. \[1=\frac{DA}{DB}.\frac{LB}{LC}.\frac{KC}{KA}=\frac{CA^2}{CB^2}.\frac{LB}{AK}.\frac{QM}{QN}\]As desired.$\blacksquare$