Right option is (c) ∑F~NB(P)\(\frac{a_F}{a_P}\) = -1
The best I can explain: From the zero sum rule,
\(a_P+\sum_{(F \sim NB(P))}a_F = 0\)
Divided by aP, the equation becomes
\(1+\frac{\sum_{F \sim NB(P)}a_F}{a_P} = 0 \)
\(\frac{\sum_{F \sim NB(P)}a_F}{a_P} =-1\)
This can be written as
\(\sum_{F \sim NB(P)}\frac{a_F}{a_P} =-1\).