Answer- Option $D$
$\text{Everyone has exactly one best friend}$
We can express it as -
$\forall x \exists y(BF(x,y) \wedge \forall z(BF(x,z) \rightarrow (y = z)))$
$\text{OR equivalently we can write it as - }$
$\forall x \exists y(BF(x,y) \wedge \forall z( (y \neq z)\rightarrow \sim BF(x,z)))$
$S1 - $ It is same as -$\forall x \exists y \forall z ((BF(x,y) \wedge \sim BF(x,z) )\rightarrow (y \neq z)) $
It is false because if some person has no best friend then LHS will be false so overall statement becomes true.
$S2-$ is false. Same reasoning as above.