Right choice is (b) ϕ^-1∈P(h)
The explanation is: Let, ϕ and σ both can fix h, then we can have ϕ(σ(h)) = ϕ(h) = h. Hence, ϕ∘σ fixes h and ϕ∘σ∈P(h). Now, all colorings can be fixed by the identity permutation. So id∈P(h) and if ϕ(h) = h then ϕ^-1(h) = ϕ^-1(ϕ(h)) = id(h) = h which implies that ϕ^-1∈P(h).