Let $AL$ meet $(ABCD)$ at $E$, so now $E$ is on $MC$ and Pascal for $BBAECD$ proves that $K-M-L$.

Let $AL//A'M$, $A'$ in $AK$. Then, $ALD$ and $A'MC$ are homothetic. Also, $\angle A'MC=\angle ALD=180-\angle ADL-\angle DAL=180-\angle ACB-\angle CAB=\angle ABC$. So, points $(A', B, C, M)$ are cyclic. Therefore, $\angle CBM=\angle CA'M=\angle DAL=\angle CAB$. In conclusion, $BM$ touches $\omega$.

Let $E = \omega \cap AL$. Then we get $BC = DE$, which means $BD \parallel CE$, hence $E, C, M$ are collinear. Now let the tangent of $\omega$ at $B$ meets $CE$ at $M'$. Applying Pascal's theorem on hexagon $BBDCEA$, we get $K, L, M'$ are collinear, which means $M = M'$, as desired. $\blacksquare$

Let $E$ be the point such that $CE\parallel BD$ and $E$ lies on circle. Then, $A,L,E$ must be collinear. Applying Pascal's Theorem on $BBDCEA$, we get that $CE\cap BB, EA\cap BD, AB\cap DC$ are collinear. We know that $L=AE\cap BD$, and $K=AB\cap DC$. So, lines $KL,CE,BB$ are concurrent. But we know that $KL,CE$ concur at $M$. Hence, $M$ lies on tangent to $B$ at circle. $\blacksquare$

okay man. okay. NEALLY. OKAY MAN. Forget $K$ exists. Let $AL$ meet the circumcircle at $X$. Define $M'$ as the intersection of the tangent from $B$ and the line parallel to $BD$ thru $C$ (which is just $XC$). We then just want to prove $M' = M$, which is equivalent to proving that $LM', AB,CD$ are concurrent. By Desargues theorem, this is equivalent to proving that $AD \cap BC, DL \cap CM', LA \cap M' B$ are collinear. We can simplify these as $AD \cap BC, BD \cap CX, AX \cap BB$. The second is just the point at infinity along $BD$, so we just want to show that the line formed by $AD \cap BC$ and $AX \cap BB$ is parallel to $BD$. We just use complex numbers, clearly $X = \frac{bd}{c}$, then $AX \cap BB = \frac{b^2(a + \frac{bd}{c}) - \frac{2ab^2d}{c}}{b^2 - \frac{abd}{c}} = \frac{2abd - abc - b^2d}{ad - bc}$. Then $AD \cap BC = \frac{adb + adc - bca - bcd}{ad - bc}$. Thus we need to verify $\frac{(adc +b^2d - bdc - adb)}{(ad - bc) (b - d)} = \frac{d(b^2 + ac - ab - cb)}{(ad - bc) (b - d)}$ is self conjugating. This is obvious.

Let $AL$ meet $\omega$ again at $X$. The result follows by pascal on $AXCDBB$.

SRMC 2022 P1 Let $AL$ intersects $\omega$ again at $Z$, then apply pascal on $BBDCZA$ and you are done.

wow Let $AL$ intersect $(ABCD)$ again at $X$. Since $\measuredangle BAX = \measuredangle CAD$. Then $BXCD$ is an isosceles trapezoid so $M$ lies on $XC$. Now pascal on $BBAXCD$ gives that the intersection of the tangent at $B$ and $XC$ lies on $KL$, however this point is $M$ proving the desired tangency.

Condition gives $\angle DAL=\angle BDC$. Note that since $MC\parallel DL$, we have $$\angle BCM=\angle BCK-\angle MCK=\angle DAB-\angle BDK=\angle LAB$$On the other hand, let point $F$ be on the segment $AK$ such that $MF\parallel AL$. Thus $\angle MFK=\angle LAB=\angle MCK$ implying $BCMF$ is a cyclic quadrilateral. And since $CF\parallel AD$, we get that $$\angle CBM=\angle CFM=\angle CFK-\angle MFK=\angle DAB-\angle LAB=\angle DAL=\angle BDC$$From which the result follows that $BM$ is tangent to $w$ at $B$.

The main claim is the following: Claim: $\triangle BMC \sim \triangle DCB$. Proof: $\angle MCB = \angle CBD$ by parallel lines, and \[\frac{CM}{BC} = \frac{CM}{DL} \cdot \frac{DL}{BC} =\frac{KC}{KD} \cdot \frac{AD}{AC} = \frac{AC}{BD} \cdot \frac{BC}{AD} \cdot \frac{AD}{AC} = \frac{BC}{BD}. \ \blacksquare\] So $\angle MBC = \angle CDB$, implying desired tangency.