Reim's theorem is useful in problems involving cyclic quadrilaterals and a line parallel to one of its sides. (Instead of using Reim's Theorem, I manually performed angle chasing to solve the problem)

Since $\angle KSB=\angle NTC = \angle NTB$ , hence $\angle NSB + \angle NTB = (180^\circ-\angle KSB ) +\angle NTB=180^\circ$.

Therefore, $B,S,N,T$ are concyclic. Now, $\angle KCB=\angle KNB=\angle SNB=\angle STB$.

So, $TS\parallel CK$.

Since $NT$ is tangent to $\omega$, hence $\angle BAN=\angle BNT$.

Also $\angle BST=\angle BNT=\angle BAN$.

$\therefore AN\parallel TS$.

Since $TS\parallel CK$ and $TS\parallel AN$, hence $CK\parallel AN\parallel TS$.