$KB$ is basically a propositional formula.
If $KB ≡ { S1, S2,\dots, Sn }$ then we can say that
$KB ≡ S1 \text{ AND } S2 \text{ AND} \dots \text{ AND } Sn$
So, $KB$ is basically a propositional formula. $S$ is also a propositional formula.
We know that $KB |= Y$ iff $(KB\rightarrow Y)$ is a tautology.
Is it possible that:
$(a)\; (KB |= S)$ and $(¬KB |= S)$
$\text{Answer:}$ Yes. For example, if $S ≡ TRUE$, i.e., if $S$ is a tautology, then any interpretation(row in the truth table) that satisfies
$KB$ or $¬KB$ also satisfies $S$.
$(b) (KB |= S) and (KB |= ¬S)$
$\text{Answer:}$ Yes. For example, if $KB ≡ FALSE$, i.e., if $KB$ is contradiction, then $KB$ entails/infers any sentence, including $S$ and $¬S$.
$\color{red}{\text{More variations:}}$
Which of the following is/are Possible ?
$\color{red}{\text{Detailed Video Solution:}}$ https://youtu.be/nclBhBmtz2g?t=685