Nice problem. Let $O$ - center of ($AKC$), then $\triangle OKC$ - equilateral, and with $BC=OC$ we have that $O$ - reflection of $B$ about $AC$. Let $L$ be reflection of $K$ about $AC$. $\triangle AKL$ - equilateral. $(*)$ $BL=KO$ like reflections of each other. So $BL=AB$. With $(*)$ it gives $\angle AKB = 150$.

This is ToT 2019 AL Junior spring round.