Note that $CM \cdot CB' = CN \cdot CA'$ and $CB' \cdot CK = CP \cdot CC' = CA' \cdot CL$, so $\frac{CM}{CK} = \frac{CN}{CL}$ and $MN \parallel KL$. Since $MK \cap NL = C$, there is a homothety centered at $C$ taking $M$ to $K$ and $N$ to $L$. This homothety must also take the midpoint of $MN$ to the midpoint of $KL$, so line $c$ must pass through $C$. Since $MNA'B'$ is cyclic, line $c$, which is the $C$-median of $\triangle CMN$, is also the $C$-symmedian of $\triangle CA'B'$. Similarly, line $a$ is the $A$-symmedian of $\triangle AB'C'$, and line $b$ is the $B$-symmedian of $\triangle BC'A'$. So $$\frac{sin\angle (AC, c)}{sin\angle (BC, c)} = \frac{sin\angle XCA'}{sin\angle B'CX} = \frac{XA'}{\frac{sin\angle CXA'}{CA'}} \div \frac{XB'}{\frac{sin\angle CXB'}{CB'}} = \frac{CA'}{CB'}$$and similarly, $\frac{sin\angle (BC, b)}{sin\angle (AB, b)} = \frac{BC'}{BA'}$ and $\frac{sin\angle (AB, a)}{sin\angle (AC, a)} = \frac{AB'}{AC'}$. Thus, $$\frac{sin\angle (AC, c)}{sin\angle (BC, c)} \cdot \frac{sin\angle (BC, b)}{sin\angle (AB, b)} \cdot \frac{sin\angle(AB, a)}{sin\angle(AC, a)} = \frac{CA'}{CB'} \cdot \frac{BC'}{BA'} \cdot \frac{AB'}{AC'} = 1$$which implies that lines $a, b$, and $c$ are concurrent via Ceva's Theorem.