Solution:
It follows from Ceva's theorem that $\frac{AM \cdot BA_{1} \cdot CB_{1}}{MB \cdot A_{1}C \cdot B_{1}A} = 1$, i.e. $\frac{CB_{1}}{B_{1}A} = \frac{CA_{1}}{A_{1}B}$. Thus, $A_{1}B_{1} \parallel AB$ and $\text{A 1 B 1 C = BAC}$. We have $\text{APM = A 1 PC = A 1 C 2 = A 1 B 1 C = BAC}$. Therefore
$\triangle PAM \sim \triangle ACM$ and analogously $\triangle PBM \sim \triangle BCM$. It follows from above that $\frac{AP}{AC} = \frac{AM}{CM} = \frac{BM}{CM} = \frac{BP}{BC}$, which implies that $\frac{AP}{BP} = \frac{AC}{BC} = \frac{AL}{BL}$, i.e. $PL$ is the inner bisector of $\text{APB}$. Moreover, $\frac{AK}{BK} = \frac{AC}{BC} = \frac{AP}{BP}$, i.e. $PK$ is the outer bisector of $\text{APB}$. Therefore $\text{LPB = 90}$ implying that $P$ lies on the circle with diameter $KL$. Note that the point $C$ lies on the same circle, which completes the proof.