I don't think that it is correct, even if you follow the example solution that explains that the given formula is equivalent to F(a∧¬b) ∨ FG¬b ∨ (Gc) ∧ (Fa), you should get another automaton. When I decompile your solution, I see
q <-> (a&!b) | Xq --> F(a&!b)
p <-> b | Xp --> F(b)
r <-> c & Xr --> G(c)
s <-> a | Xs --> F(a)
However, instead of F(b), there should be FGb which is not the same. Why do you believe that it is the same?