$\angle AOC=2\angle B=60$ so $ACO$ is equilateral triangle. $\angle KAO=120, \angle KLO=60$ Let $\angle ALK=x$ then $\angle OCL=\angle COL=\angle CKL=\angle LKO=30-x, \angle AOK= 60-\angle AKO=2x, \angle AOL=90-x=\frac{\angle AOB}{2}=\angle LOB$ So $\triangle AOL=\triangle BOL $ and $\angle LBO=\angle LAO=30$ $\angle KAL+\angle LBK=180$ so $A,B,L,K$ are concyclic.

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Remark: If $E$ is the circumcenter of $(KOC)$, then $O,E,K,A$ are concyclic also. Best regards, sunken rock