The problem can be solved by angle chasing, by algebraic equations resulting from similar triangles, or by a combination of the two methods.

We first show that $S, A, T$ are collinear. Because $ASBK$ is cyclic, we have $\angle SAB = \angle SKB$. From $KS \parallel AC$ we get $\angle SKB = \angle C$, so $\angle SAB = \angle C$. Similarly $\angle TAC = \angle B$, which shows that $S, A, T$ are collinear as
$$\angle SAT = \angle A + \angle B + \angle C = 180^\circ.$$

Quadrilateral $SLMT$ is cyclic iff $\angle KLM = \angle ATM$. But $\angle ATM = \angle C$ and $\angle KLM = \angle AML$ so $MLST$ is cyclic if and only if $\angle AML = \angle C$, which is equivalent to $ML \parallel BC$.

We will show in two ways that $ML \parallel BC$ is equivalent to $K$ being the midpoint of $BC$.

Version 1. Suppose $ML \parallel BC$. Because $KM \parallel AB$ and $KL \parallel AC$, we have two parallelograms $BKML$ and $CKLM$ hence $BK = ML = CK$.
Reciprocally, if $K$ is the midpoint of $BC$, then $KM \parallel AB$ and $KL \parallel AC$ imply that $L$ and $M$ are the mid-points of $AB$ and $AC$, respectively. Hence $ML$ is a midline in $\triangle ABC$ and so $ML \parallel BC$.

Version 2. By the Intercept Theorem, $ML \parallel BC$ is equivalent to $\frac{AM}{MC} = \frac{AL}{LB}$. On the other hand, $KL \parallel AC$ and $MK \parallel AB$ imply
$$\frac{CK}{KB} = \frac{AL}{LB} \quad \text{and} \quad \frac{AM}{MC} = \frac{BK}{KC}.$$
Hence, $SLMT$ is cyclic if and only if $\frac{CK}{KB} = \frac{BK}{KC}$, i.e. $CK = KB$.

Solution 2. (based on a strategy by AngYang Li)

Denote lengths of segments as follows:
$$\begin{aligned} c &= |AB| & x &= |AM| = |KL| \\ b &= |AC| & y &= |AL| = |MK|. \end{aligned}$$
Since $LK \parallel AC$ and $MK \parallel AB$, we have $\triangle MKC \sim \triangle ABC \sim \triangle LBK$, hence
$$\frac{b-x}{y} = \frac{b}{c} = \frac{x}{c-y} \quad \text{and so} \quad y = \frac{c}{b}(b-x) \quad \text{and} \quad c-y = \frac{c}{b} \cdot x. \quad (13)$$

$$\begin{aligned} |KM| \cdot |MT| &= |AM| \cdot |MC| = x(b-x) \\ |KL| \cdot |LS| &= |AL| \cdot |LB| = y(c-y). \end{aligned}$$
Finally, $SLMT$ is cyclic iff $\triangle KLM \sim \triangle KTS$, which in turn is equivalent to $\frac{|KM|}{|KL|} = \frac{|KS|}{|KT|}$ which we rewrite as follows in equivalent ways
$$\begin{aligned} & |KM| \cdot |KT| = |KL| \cdot |KS| \\ & |KM|^2 + |KM| \cdot |MT| = |KL|^2 + |KL| \cdot |LS| \\ & y^2 + x(b-x) = x^2 + y \cdot (c-y) \\ \end{aligned}$$
$$(b - 2x) \left( \frac{c^2}{b^2} (b - x) + x \right) = 0.$$
The last equation holds true iff $b = 2x$, i.e. $M$ is the midpoint of $AC$, which is equivalent to $K$ being the midpoint of $BC$, as $MK \parallel AB$.