Let $\{X\}=(KNS)\cap (ABC)$. From a simple angle chasing, $MBXK$ is cyclic, hence $\{X\}=(BMK)\cap (BAC)$, i.e. $X$ is the Miquel point of the complete quadrilateral $ABCMKL$. It is well known that $X$ is the center of the spiral similarity which takes $BC$ to $ML$. The same spiral similarity takes

$K$ to $N$, hence $\angle (XK,XN)=\angle (XB,XM)$. Thereby, $\widehat{KSN}=\widehat{KXN}=\widehat{BXM}=\widehat{BKM}=\widehat{LKC}$ so the claim is proved.

My solution: Let $ X=AN \cap BC $ and $ Y $ be a point on $ BC $ such that $ AY \parallel KL $ . Since $ (Y,X;B,C)=(AY,AN;AM,AL)=-1 $ , so $ XA \cdot XS=XB \cdot XC=XK \cdot XY \Longrightarrow A, S, K, Y $ are concyclic , hence we get $ \angle NSK=\angle ASK=\angle AYK=\angle NKX \Longrightarrow BC $ is tangent to $ \odot (NSK) $ at $ K $ . Q.E.D

We have $$\frac 12 = \frac{LN}{LM}= \frac{AN \cdot \sin \angle LAN}{AM \cdot \sin \angle LAM} = \frac{AN}{AM} \cdot \frac{\sin \angle CBS}{\sin \angle BSC}= \frac{AN}{AM} \cdot \frac{SC}{BC}.$$ Therefore $\frac{SC}{AM}= \frac{BC}{2AN}= \frac{KC}{AN}$. This follows $\triangle CSK \sim \triangle AMN \; ( \text{S.A.S})$. Hence, $\angle ANM= \angle CKS$. This leads to $\odot (KNS)$ is tangent to $BC$.

consider the triangle $(AMN) $ and the points $B,S $ and $ K $ are on $AM,AN$ and $ MN$ respectively.Miquel theorem assure that $ (ABS),(KSN) $and$(MBK)$ have a commun point say $P$.Idem if we consider $(AML) $ and the points $B,C $ and $ K $ are on $AM,AL$ and $ ML$ respectively.Miquel theorem assure $ (ABC),(KLC) $and$(MBK)$ have a commun point which is $ P$ then $ (KSN),(KLC) $ and $MBK$ are coaxal thus the ratio of the powers wrt $ (KSN)$ and $(KLC) $ are the same for $M$ and $B$ which leads to the power of $B $ wrt $(KSN)$ is $BK^2$ ie $(KSN)$ is tangent to $BC$

Let $P$ be the reflection of $A$ across $N$. Note $AMPL$ is a parallelogram. Then $\odot KSN$ tangent to $BC$ $\Leftrightarrow \angle{BKS}=\angle{KNS}=\angle{ANL}$ $\Leftrightarrow$ $\triangle BKS\sim \triangle ANL$ $\Leftrightarrow$ $\triangle BCS\sim \triangle APL$. This is true since $\angle{PAC}=\angle{SBC}$, and $\angle{SCB}=\angle{SAB}=\angle{APL}$.

$NL/ML = \frac{AN \cdot \sin \angle LAN}{AM \cdot \sin \angle LAM} = AN/AM \cdot SC/BC$ so $AM/AN = 2CS/CB = CS/CK,$ $\angle KCS = \angle MAN$ so $MAN$ and $CSK$ are similar. Let $SNK$ meet $BS$ at $F$. $\angle ANM = \angle KFS = \angle SKC$ so $SKN$ is tangent to $BC$. we're Done.

Let $P$ be the Miquel point of $AMBKLC.$ Since $$\measuredangle NSP=\measuredangle ABP=\measuredangle MKP=\measuredangle PKN,$$$PKNS$ is cyclic. Since $P$ is the center of the spiral similarity $\overline{BC}\mapsto\overline{ML},$ it is the center of $\overline{BK}\mapsto\overline{MN}.$ Hence, $\triangle KPB\sim\triangle NPM$ and $\measuredangle BKP=\measuredangle KNP.$ $\square$