Proof (a).

\$$K_B\$$ is the claimed constant functor which maps every set to \$$B\$$ and every morphism to \$$Id_B\$$.

To prove that \$$K_B\$$ is indeed a functor, it suffices to show that for all sets \$$S\$$, the mapping \$$K_S\$$ satisfies the two functor laws.

The identity law for \$$K_S: Set \rightarrow Set\$$ says that for every \$$A \in Ob(Set)\$$, we have that \$$K_S(Id_A) = Id_{K_S(A)}\$$. Since \$$K_S(A) = S\$$, this requirement restates to \$$K_S(Id_A) = Id_S\$$.
This is certainly true, as \$$K_S\$$ maps any morphism whatsoever to \$$Id_S\$$.

The composition law requires that for composable morphisms \$$f, g\$$, we have that \$$K_S(f \triangleright g) = K_S(f) \triangleright K_S(g)\$$. That's equivalent to saying that \$$Id_S = Id_S \triangleright Id_S\$$, which is an axiomatic fact.