On a non-empty set $A,$ we define a relation $S.$
$∀x∀y(S(x,y)⇔¬S(y,x))$ means that for all pairs $(x,y) \in A \times A,$ $S(x,y)⇔¬S(y,x)$ must be true, which is NEVER possible to be true for any non-empty set $A$ because when $x,y$ refer to same element, $S(x,y)⇔¬S(y,x)$ will be False.
Hence, there is NO finite Or infinite model for the given expression, Or we can say that on any non-empty set $A,$ we cannot define any such relation $S.$
Hence, answer is C,D.
In the same question, if we have the following definition of relation $S:$
$∀x∀y(S(x,y) \rightarrow ¬S(y,x))$
∧
$[∀x∃yS(x,y)]$
∧
$∀x∀y∀z(S(x,y)∧S(y,z)⇒S(x,z))$
Now, it becomes satisfiable but has no finite model.
Similar question and my explanation you can find here: GATE CSE 1991 | Question: 15,b Deepak Poonia Explanation