Well, let's break it up into easy pieces...
You've kept a visited array, and by looking it up, you decide whether to push a node into the queue or not. Once visited, you don't push it again. So, how many nodes get pushed into the queue: (of course) V nodes. And it's complexity is O(V).
Now, each time, you take out a node from queue and visit all of its neighboring nodes. Now, following this way, for all of V nodes, how many node you'll come across. Well, it's the number of edges if the graph is unidirectional, or, 2 * number of edges if the graph is bidirectional. So, the complexity would be O(E) for unidirectional and O(2 * E) for bidirectional.
So, the ultimate(i.e. total) complexity would be O(V + E) or O(V + 2 * E) or generally, we may say O(v + E).
