The keyboardist count is capped at one, so split the count on that value; the strictly-more-guitarists-than-drummers condition and the at-least-one-drummer condition then fix which splits are allowed within each case.
Zero keyboardists: the 5 members come from guitarists and drummers, with more guitarists than drummers and at least one drummer. The only splits that satisfy both conditions are 4 guitarists with 1 drummer, C(5,4) × C(4,1) = 5 × 4 = 20, and 3 guitarists with 2 drummers, C(5,3) × C(4,2) = 10 × 6 = 60. This case totals 80.
Exactly one keyboardist: choose the keyboardist in C(3,1) = 3 ways, then fill the remaining 4 seats from guitarists and drummers under the same two conditions. The only valid split is 3 guitarists with 1 drummer, since a 4-and-0 split has no drummer and a 2-and-2 split is not strictly more guitarists than drummers: C(5,3) × C(4,1) = 10 × 4 = 40, and the 3 choices for the keyboardist give 3 × 40 = 120.
Adding the two cases gives 80 + 120 = 200. So the answer is D.