Solved with mueller.25, AdhityaMV, Siddharth03. Let $O$ be the circumcenter of $\triangle BCK$. Note that $P$ is just the reflection of $O$ over $BK$. Reflect $B$ over $O$ to get the antipode $B'$, which clearly lies on $AC$. Since $BKB'C$ is cyclic, we have $\triangle ABB' \sim \triangle ACK$. Since $L$ and $O$ are midpoints, we also get that $\triangle ABP \sim \triangle ABO \sim \triangle ACL$. So $A$ is the center of spiral similarity sending $CL$ to $BP$, and so also sends $BC$ to $PL$. So $\angle APL = \angle ABC = \angle PBL$, so $AP$ is indeed tangent to the circumcircle of $BLP$, as desired. $\blacksquare$