If $L, M, N, P$ are the tangency points of $AB, CD, DE, EA$ respectively with the incircle of center $I$, then $\measuredangle BIK=\measuredangle DIM=\measuredangle DIN=\measuredangle BIL=\alpha$ and $\measuredangle AIL=\measuredangle AIP=\measuredangle CIK=\measuredangle CIM=\beta$, then $\measuredangle KIA =\measuredangle KIN =2(\alpha+\beta)$. As $E$ lies on the angle bisector of $\angle PIN$, the result follows. Best regards, sunken rock

It must be $$KIP = KIN$$

Let the tangent points of the circle to the pentagon be $J$ to $AB$, $K$ to $BC$, $L$ to $CD$, $M$ to $DE$, and $N$ to $EA$. Then \[AJ+JB=BK+KC\rightarrow NA=AJ=KC=CL.\]Similarly, \[JB=BK=LD=DM.\]Therefore, \[\triangle BKS\cong \triangle BJS\cong \triangle DLS\cong \triangle DMS\]by SAS. So $\angle ABK=\angle CDM$, and similarly $\angle NAB=\angle KCD$. So quadrilaterals $NABK$ and $KCDM$ are congruent, meaning that $KM=KN$. Since $SM=SN$ and $EM=EN$ as well, $S,E,K$ are all on the perpendicular bisector of $MN$. $\blacksquare$