Turning the amino group into an amide serves two purposes:
We are introducing a sterically bulky group. We expect the amide bond to be as complanar as possible with respect to the benzene ring to maximise electronic interactions. That should block one of the two ortho positions sterically.
We are turning an electron-rich, $+M$ aromatic system into an electron-poor, $-I$ one.
Especially the second transformation will greatly reduce the reaction rate. Furthermore, a bromine atom is mildly deactivating, too, due to its $-I$ effect (the weak $+M$ effect is neglegible and only responsible for the ortho/para directing abilities). Thus, it is very likely that we can selectively stop the reaction after monobromination.
The question still arises why we only obtain the para-product. Obviously the amide behaves much like a bromide and displays a weak $+M$ effect in spite of the amide resonance. Yet we can imagine the the proximity to the bulky amide group inhibits substitution at both ortho-protons sufficiently to allow isolation of the para-product.