redefine M to be center of (MAL) and gg after some chase

From the tangency and length condition we have \[ \angle KAB=\angle BAC=\angle CAD=\angle BCK=\angle BDC=\angle BDA=\angle BMA=\angle BCA=\angle DCL=\angle DLC \]That gives $\angle AMK=\angle ALK=\angle AMB$, means $K, B, M$ are collinear. Also notice that $AKCB$ is a kite because of $KA=KC$, $BA=BC$, $\angle KAB=\angle KCB$. So $KM$ bisects $\angle AKL$.

Define $N$ as the antipode of $B$ wrt $\omega$. We will show: $N \ne A$. $N \in \odot(AKL)$ $NA = NL$. This will show $M \equiv N$ and $MA = ML$. Its clear $N \ne A$, otherwise $C \equiv A$, contradiction. Now $AB = CD$ gives $BC \parallel AD$. As $BC \perp CN$, thus $$ CN \perp AD $$Now note points $K,B,N$ are collinear on the perpendicular bisector of segment $AC$. This also means $$ KN \text{ bisects } \angle AKC \qquad \qquad (1) $$[asy][asy] size(200); pair O=(0,0),B=dir(117),C=2*foot(B,O,dir(90))-B,A=2*foot(C,O,B)-C,D=2*foot(B,O,C)-B,K=2*A*C/(A+C),L=2*foot(C,A,D)-A,N=-B; fill(B--A--C--cycle,green+white+white); fill(C--D--L--cycle,green+white+white); draw(unitcircle,royalblue); draw(circumcircle(K,A,L),purple); draw(A--K--L,brown); draw(B--C^^A--L,orange); draw(N--C,red); draw(N--K); draw(A--N--L,cyan); draw(B--A--C--D,magenta); dot("$A$",A,dir(A)); dot("$B$",B,dir(160)); dot("$C$",C,dir(C)); dot("$D$",D,dir(-40)); dot("$K$",K,dir(K)); dot("$L$",L,dir(L)); dot("$N$",N,dir(N)); markscalefactor=0.015; draw(anglemark(A,K,B)); markscalefactor=0.018; draw(anglemark(N,K,C)); [/asy][/asy] Now in $\triangle CBA$ and $\triangle CDL$ we have: $$ CB = CD ~,~ \angle CBA = \angle CDL ~,~ \angle DCL = \angle BCA $$It follows $$ \triangle CBA \cong \triangle CDL $$Thus $CA = CL$. As $CN \perp AL$, so it follows $CN$ is perpendicular bisector of segment $AL$. Hence, $$ NA = NL $$Look at $\triangle AKL$ now. Note $$ KL > KC = KA $$particularly $KA \ne KL$. Now $N$ is the intersection of perpendicular of $AL$ and internal angle bisector of $\angle AKL$. As $KA \ne KL$, it follows $N$ is the midpoint of arc $\widehat{AL}$ of $\odot(AKL)$ not containing $L$. This completes the proof. $\blacksquare$

As $AB = BC = CD$ we have that $ABCD$ is a isosceles trapezium. We denote $\angle ACB = \angle BAC = \angle CAD = \alpha$. Also $\angle CDL = 180^{\circ} - \angle CDB - \angle BDA = 180^{\circ} - \angle BAC - \angle ACB = 180 - 2\alpha.$ As $\angle LCD = \angle CAD = \alpha \Rightarrow \angle CLD = \alpha$. Now, $\angle KAB = \angle BDA = \alpha$ and $\angle KCB = \angle CDB = \alpha$ and as $\triangle ABC$ is isosceles we have that $KB$ is perpendicular bisector of $AC$. By the cyclic $KLMA$ $\angle KMA = \angle KLA = \alpha$, but $\angle BMA = \angle BDA = \alpha = \angle KMA$, then $K$, $B$ and $M$ are collinear. Finally, $\angle MAL = \angle MKL = \angle BKC = \angle BKA = \angle MKA = \angle MLA $ thus $MA = ML$, as desired $\blacksquare$

Attachments:

By angles chasing we have $\angle KAB=\angle KCB=\angle KLA$. Invert the point $L$ via the circle $\mathcal{C}\left(K,KA\right)$ to be $L'$. Note that $\angle KAL'=\angle KLA=\angle KAB$, so we have $A,B,L'$ are collinear. Then we invert this line to be the circumcircle of $\triangle KLA$ which meets $\omega$ again at $B'$, the image of $B$ over the inversion. This implies $M=B'$ and hence $M,B,K$ are collinear. Hence $KM$ bisects $\angle AKL$, so we are done.

just bash it set $\omega$ as the unit circle and let $a=1,b=x$ so $c=x^2 ,d=x^3$ and let $M'$ be the antipode of $B \implies m'=-b=-x$ $k=\frac{2ac}{a+c}=\frac{2x^2}{x^2+1} ,l=\frac{cc(a+d)-ad(c+c)}{cc-ad}=\frac{x^7+x^4-2x^5}{x^4-x^3}=x^3+x^2-x$ now let's prove that $M'AKL$ is cyclic : $$\Delta=\frac{k-a}{k-m'}/\frac{l-a}{l-m'}=\frac{(x^2-1)(x^2+x)}{(x^2+2x+1)(x^3+x^2-x-1)}=\frac{x}{(x+1)^2}=\frac{\frac{1}{x}}{(\frac{1}{x}+1)^2}=\overline{\Delta} \implies M'=M$$$\implies ML=|x^3+x^2-x+x|=|x^3+x^2|=|x^2|.|x+1|=|x+1|=MA$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $$ \blacksquare$

Use complex numbers with $\omega =\mathbb{S}^1$. Set $A=1$ and $B=b$, then $C=b^2$ and $D=b^3$. Define $M'=-b$. We have $K=\frac{2AC}{A+C}=\frac{2b^2}{b^2+1}$ and\begin{align*}L & =\frac{2ACD-(A+D)C^2}{AD-C^2} \\ & =\frac{2b^5-(b^3+1)b^4}{b^3-b^4} \\ & =b(b^2+b-1). \end{align*}Hence\begin{align*}\frac{K-A}{L-A}\frac{L-M'}{K-M'} & =\frac{\frac{2b^2}{b^2+1}-1}{b(b^2+b-1)-1}\frac{b(b^2+b-1)-(-b)}{\frac{2b^2}{b^2+1}-(-b)} \\ & =\frac{b}{(b+1)^2}\in \mathbb{R}. \end{align*}Therefore $M'=M$. Now clearly $K,B,O,M$ are collinear, so $KM$ is the angle bisector of $\angle AKC$, which gives $MA=ML$, as wanted.

Let: $O:\text{center of} \omega$, so $O\in AD$ Let: $C':\text{ antipode of } C$ $\angle KAL=\angle KML=90^{\circ}$ Now $\angle CLM = \alpha=180^{\circ}-\angle KAM=\angle MCA$ But $\angle BCA=\angle KCB=\beta, \text{ but } \angle CMD=\beta\Rightarrow\angle DML=\beta$, so $\triangle CDL\sim \triangle CBA$ so $KB\perp ML$ But $\angle LKB=\angle AKB=\beta \text{ and } KAML \text{ is cyclic}$ So $M\in \text{ bisector of } LA$, therefore $MA=ML$, As desired $ \hspace{20cm} \blacksquare$