Decidability of a set containing first symbols of elements of a decidable set
You probably have to twist your head a couple of times but $B$ is decidable and it does not depend on the decidability of $A$. Note that $B \subseteq \Sigma$, i.e. a subset of your finite alphabet. Thus, $B$ is finite and hence, it is decidable.

However, this reasoning is not constructive: It is just enumerating all functions $\Sigma \to \{0, 1\}$ (that are $2^{|\Sigma|}$, i.e. finitely many), which are all computable, and one of them is the correct function although you do not know which one.