Let $ P \equiv OK \cap BC.$ From the cyclic quadrilaterals $ ABKO,DCKO$ it's easy to see that $ KP$ bisects $ \angle BKC$ since $ \angle BKP = \angle BAO = \angle ODC = \angle CKP.$ Let the angle bisectors of $ \angle KBL$ and $ \angle KCL$ cut $ LC,LB$ at $ B',C'.$ $ \angle KCL = \angle KCB + \angle ADK = \angle ACB - \angle KCO + \angle ADB + \angle ODK$ $ \Longrightarrow \angle KCL = 2\angle BCA = \angle BCC' = \angle KCA$ Similarly, $ \angle CBB' = \angle KBD,$ which implies that $ CA,CC'$ and $ BD,BB'$ are isogonals WRT $ \angle KCB$ and $ \angle KBC,$ respectively. Hence if $ I \equiv BB' \cap CC',$ then $ O,I$ are isogonal conjugates WRT $\triangle BKC,$ but angle bisector $ KO$ of $ \angle BKC$ is self-isogonal $ \Longrightarrow$ $ I \in KP$ $\Longrightarrow$ internal angle bisectors of the convex quadrilateral $ BLCK$ concur at $I.$