We will show that $\cot\angle LKM = \cot\angle MBL$. Note that $$\cot\angle MBA =\frac{\frac{BA}{AM}-\cos\alpha}{\sin\alpha}=\frac{\frac{2c}{b}-\cos\alpha}{\sin\alpha}.$$Since $\angle MBA+\angle MBL =180^{\circ},$ we obtain that $$\cot\angle MBL = -\cot\angle MBA=\frac{\cos\alpha-\frac{2c}{b}}{\sin\alpha}.$$Law of Sines in $\triangle AML$ gives us that $$\frac{ML}{\sin\alpha}=\frac{MA}{\sin(90^{\circ}-\alpha-\gamma)}\Rightarrow ML=\frac{\frac{b}{2}\sin\alpha}{\cos(\alpha+\gamma)}.$$Also, we have that $KM=MD=AM-MD=\frac{b}{2}-c.\cos\alpha.$ Hence, we have that $$\cot\angle LKM = \frac{\frac{KM}{ML}-\cos\angle KML}{\sin\angle KML}=\frac{\frac{(\frac{b}{2}-c.\cos\alpha)\cos(\alpha+\gamma)}{\frac{b}{2}\sin\alpha}-\sin\gamma}{\cos\gamma}.$$Now $\cot\angle LKM = \cot\angle MBL\iff$ $$\iff \frac{\cos\alpha-\frac{2c}{b}}{\sin\alpha} = \frac{(\frac{b}{2}-c.\cos\alpha)\cos(\alpha+\gamma)-\frac{b}{2}\sin\alpha\sin\gamma}{\frac{b}{2}\sin\alpha\cos\gamma}\iff$$$$\iff\frac{b}{2}(\cos\alpha\cos\gamma-\cos(\alpha+\gamma)+\sin\alpha\sin\gamma)=c(\cos\gamma-\cos\alpha\cos(\alpha+\gamma))\iff$$$$\iff\frac{b}{2}\cdot 2\sin\alpha\sin\gamma=c(\cos\alpha\cos\beta-\cos(\alpha+\beta))\iff$$$$\iff b\sin\alpha\sin\gamma = c\sin\alpha\sin\beta\iff$$$$\iff \frac{b}{\sin\beta}=\frac{c}{\sin\gamma},$$which is true, so we are done.

Cute. We define some points, $B'$ be the parallelogram point, Let the circle with diameter $CB'$ intersect $AC$ at $B_2'$, $BB'$ at $G$, $BM$ at $F$. Let $E$ be on $AC$ such that $ME \perp AC$. Claim 1: $ABKF,ABB_2'G$ is cyclic

Proof

Redefine $L = AB \cap FK \cap B_2'G$ we claim $MKB_2'L$ cyclic which would finish as $B$ and $B_2'$ are reflection WRT $ME$. Firstly it is easy to see that $ML \perp AC$ which clearly finishes as $\measuredangle KB_2'L = \measuredangle ABK = \measuredangle KML$ so done