You do know that in the standard order or operations division comes before addition, right?

It's not wrong. First, since B and C are mutually exclusive (i.e., are disjoint subsets of the sample space), we have P(B) + P(C) = P(B ∪ C). Second, for the same reason A ∩ B and A ∩ C are also disjoint, so P(A ∩ B) + P(A ∩ C) = P((A ∩ B) ∪ (A ∩ C)) = P(A ∩ (B ∪ C)).