As VicKmath7 pointed out, this problem was created by coincidence a few days before EGMO Day 2 happened lol . Simple angle chasing shows that $AQKC$ is cyclic. Now if we consider the negative inversion which preserves its circle, then we get as equivalent to show $TM \parallel X'Y'$, where $X'$ and $Y'$ are the intersections of the prolongations of the sides of $AQKC$. But by Brocard this becomes equivalent to $OM \perp MT$. For this, let $T' = MT \cap AK$, we have $MT = TT'$ by the parallelogram $AKBP$ and we conclude by the butterfly theorem in $AQKC$.

Two remarks: - The part from $TM \parallel X'Y'$ can be proven by the method of moving points. - Can this problem be nuked with Desargues Involution Theorem?

Complex numbers FTW! First, a few synthetic remarks: $MA=MB$ and $MP=MK$, so $AKBP$ is a parallelogram. Now notice that by a quick angle chase $AQKC$ is cyclic: $\angle KCQ=\angle PCQ=\angle PBQ=\angle QAK\Longrightarrow AQKC$ is cyclic. Now notice that by angle chasing using the given cyclic quadrilaterals $AKMX$, $BCXM$, $AMYC$, $KMYB$ we get: \[\angle XAB=\angle XAM=\angle XKM=\angle XKC,\quad \angle XBA=\angle XBM=\angle XCM=\angle XCK\Longrightarrow \triangle XAB\sim\triangle XKC\]\[\angle YAB=\angle YAM=\angle YCM=\angle YCK,\quad \angle YBA=\angle YBM=\angle YKM=\angle YKC\Longrightarrow \triangle YAB\sim \triangle YCK\]And now we can complex bash: Let $(AKQC)$ be the unit circle. Then, by using the center of a spiral similarity formula and the intersection of chords in the unit circle formula we have that: \[x=\frac{ac-kq}{a+c-k-q},y=\frac{ak-cq}{a+k-c-q}, m=\frac{aq(k+c)-kc(a+q)}{aq-kc}\]Thus we only need to calculate $t:$ $$T\in CQ\Longrightarrow t=c+q-cq\overline{t}$$$$T\in BP\Longrightarrow BT\parallel AK\Longrightarrow \frac{b-t}{a-k}=\overline{\left(\frac{b-t}{a-k}\right)}=\frac{ak(\overline{b}-\overline{t})}{k-a}$$$$\Longrightarrow -2m+a+t=ak\left(2\overline{m}-\frac{1}{a}-\frac{c+q-t}{cq}\right)$$$$\Longrightarrow -2mcq+acq+tcq=2\overline{m}akcq-kcq-ak(c+q-t)$$\[\Longrightarrow t=\frac{2mcq+2\overline{m}akcq-acq-kcq-akc-akq}{cq-ak}\]Now $TM$ is tangent to $\odot\triangle MXY\iff\angle TMY=\angle MXY$ (we're working with oriented angles here) \[\iff \frac{t-m}{y-m}\Big/\frac{m-x}{y-x}=\overline{\left(\frac{t-m}{y-m}\Big/\frac{m-x}{y-x}\right)}\iff \frac{(t-m)(y-x)}{(y-m)(m-x)}\in \mathbb{R}\]However, we have: $$m-x=\frac{aq(k+c)-kc(a+q)}{aq-kc}-\frac{ac-kq}{a+c-k-q}=\frac{(a-k)(c-q)(cq-ak)}{(aq-kc)(a+c-k-q)}$$$$y-m=\frac{ak-cq}{a+k-c-q}-\frac{aq(k+c)-kc(a+q)}{aq-kc}=\frac{(a-c)(k-q)(ac-kq)}{(aq-kc)(a+k-c-q)}$$$$y-x=\frac{ak-cq}{a+k-c-q}-\frac{ac-kq}{a+c-k-q}=\frac{(a-q)(k-c)(a+q-c-k)}{(a+c-k-q)(a+k-c-q)}$$$$t-m=\frac{2mcq+2\overline{m}akcq-acq-kcq-akc-akq}{cq-ak}-m=$$$$=\frac{mcq+mak+2\overline{m}akcq-acq-kcq-akc-akq}{cq-ak}$$By substituting $m=\frac{aq(k+c)-kc(a+q)}{aq-kc}$ and $\overline{m}=\frac{c+k-a-q}{kc-aq}$ into the expression above we get:

), so we're done!

Here is fully the nicest in my opinion (in terms of technical details) solution which we are aware of. Clearly $AKBP$ is a parallelogram and so with the cyclic $BQPC$ we get $\angle AKC = \angle KPB = \angle AQC$, i.e. $AKQC$ is cyclic - denote its center by $O$. Consider the composition $f$ of inversion centered at $M$ with radius $\sqrt{MA \cdot MQ} = \sqrt{MC \cdot MK}$ and reflection with respect to $M$. Denote $AK \cap CQ = X'$ and $AC \cap KQ = Y'$. Then $f(A) = Q$, $f(C) = K$ and inversion properties give $f(X) = X'$ and $f(Y) = Y'$. Since the line $TM$ (which passes through the center $M$ of inversion and reflection) is mapped to itself, the desired statement is equivalent to $TM \parallel X'Y'$. Now Brocard's theorem for the cyclic quadrilateral $AKQC$ gives $OM \perp X'Y'$ and so it is now enough to show $OM \perp MT$. If $TM \cap AK = T'$, then $BT = AT'$ and $T'K = TP$ from the parallelogram $AKBP$. Let $TM$ intersect the circle of $AKQC$ at $X_0$ and $Y_0$, so that $T$ is between $M$ and $Y_0$ - then $X_0T' + T'Y_0 = X_0Y_0 = X_0T + TY_0$ and $X_0T' \cdot T'Y_0 = AT' \cdot T'K = BT \cdot TP = CT \cdot TQ = X_0T \cdot TY_0$ and since $X_0T' < X_0T$, we get $X_0T' = TY_0$. Together with $MT = MT'$ from the parallelogram $AKBP$ we get $X_0M = MY_0$ and so $OM \perp X_0Y_0 \equiv MT$, as desired

Inversion + Brocard thrm. + power of a point.

We see that $MC.MK=MC.MP=MB.MQ=MA.MQ$ thus, $A,C,Q,K$ are concyclic. Invert at $M$ with radius $MA$. New Problem Statement: $QCAK$ is a cyclic quadrilateral and $AK\cap CQ=X,AC\cap QK=Y,AQ\cap CK=M$. $B,P$ are the reflections of $A,K$ with respect to $M$. If $(MCQ)\cap (MBP)=T$, then prove that $MT\parallel XY$. Let $BP\cap CQ=R$. Since $B,C,P,Q$ are concyclic, $M,R,T$ are collinear. Let the perpendicular to $OM$ at $M$ intersect $QC,AK$ at $U,V$ and $(ACQK)$ at $U_1,V_1$. Since $MU_1=MV_1$, by Butterfly theorem we get $MU=MV$ hence $BU\parallel AV\equiv AK$. Combining this with $AK\parallel BR$ implies $R=U$. Since $MR\perp OM\perp XY$ by Brocard, we get the desired result.$\blacksquare$