Because you actually have one mercury cation and two bromide anions.
The thing to remember about mercury(I) bromide,
Mercury is still in its +1 oxidation state, it's just that two such cations form a metal-metal bond and exist as
This means that the compound is formed using two bromide anions,
These mercury cations are actually called polycations because they feature two mercury(I) cations bonded together.
This means that when mercury(I) bromide is placed in aqueous solution, it's dissociation equilibrium will be
#"Hg"_2"Br"_text(2(s]) rightleftharpoons "Hg"_text(2(aq])^(2+) + color(red)(2)"Br"_text((aq])^(-)#
By definition, the solubility product constant,
#K_(sp) = ["Hg"_2^(2+)] * ["Br"^(-)]^color(red)(2)#