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Statically determinate. A pin gives two reaction components and a roller one, so there are three support unknowns and nine member unknowns, twelve in all. Each of the six joints supplies two equilibrium equations, so there are twelve equations — and m = 2j − 3 reads 9 = 12 − 3 ✓. Option D confuses two independent things: determinacy is a property of the structure, and the loading never enters it. The count is of members, joints and reactions only, which is why you can settle it before reading what the loads are. Option B would need a tenth member or a second roller; option C a member removed, after which some joint could move without stretching anything and no equilibrium solution would exist. One caveat the count cannot catch: a truss can satisfy m = 2j − 3 and still be unstable if the members are badly arranged — three collinear members at a joint, say — so the count is necessary and not sufficient.