From group theory we know that a homomorphism $\phi: G \to \operatorname{Sym}(S)$, where S is a set, then $\operatorname{Sym}(S) \cong \Sigma_n $. Its kernel is given as $\bigcap_{s \in S}G_s$, which is the intersection of all the stabilizer subgroups of $G$.
Now, the pure braid group $P_n$ is defined as the kernel of the map $\phi: B_n \to \Sigma_n$. Is it fair to say, that the pure braid group is such an intersection?
Assuming my comment is correct, there is no restriction in the hypothesis on the group $G$. The statement is true for any $G$ at all. In particular it is true for $G=B_n$.