(assuming $L\not= B$) $\angle KMB=\angle KCB=90^{\circ}$ so triangle $KAB$ is isosceles. It follows that $MLC$ is the orthic triangle of triangle $KAB$ so $AL$ and $KM$ intersect at it's orthocentre, from which the result follows.

Since $KO\cap \odot(ABC)=L,$ therefore we have $\angle ALB=\dfrac{\pi}{2},$ leading to $A,L,K,M$ being cyclic. Now, applying the Radical Axes theorem on $\odot(AMLK),\odot(ABC),$ and $\odot(KCMB),$ we observe that the lines $AL,KM,BC$ concur at the radical centre of these circles. Let this point of concurrence be $H.$ Since $\angle HCA=\dfrac{\pi}{2}=\angle HMC,$ therefore $HMAC$ is a cyclic quadrilateral. In other words, $H\in\odot(AMC),$ as desired. $\Box$

Attachments:

Without loss of generality we assume $\angle B>45^{0}$,which implies that the circumcentre P of $\triangle AMC$ lies outside it. Now $\angle PAC=\angle AMC-90^{0}$ Also $\angle LAC=\angle KBC=90^{0}-\angle BMC=\angle AMC-90^{0}$ which implies that $P$ lies on $AL$,i.e $AL$ is a part of a diameter of circle $AMC$,also $\angle AMK=\angle ACB=90^{0}$ which implies that $MK$ and $AL$ intersect on the circumcircle of $\triangle AMC$.

Assume $CA<CB$, the other case being fairly identical. We show that $\triangle CMA \sim \triangle CKL$. Indeed, we see that $\angle CKL=90^{\circ}-\angle CBK$ and since $\angle CBK=\angle CMK=180^{\circ}-2\cdot \angle CBA-90^{\circ}$ we get that $\angle CKL=\angle CMA=2\cdot \angle CBA$. Now, $\angle CKM=\angle CBA$ and so $KM$ is an angle bisector of $\angle CKL$. This yields that $\triangle CKL$ is isosceles since $MC=ML$. Thus, by the spiral similarity lemma, $KM \cap AL$ lies on $(CMA)$. The result holds.