Let k be the order of a mod n then a^b ≡ 1(mod n) if and only if 

Concept:

Let b ∈ Z such that a^b ≡ 1(mod n).

Let us apply division algorithm to b and k then we have,

b = kq + r where 0 ≤ r ≤ k

Consider, a^b = a^{kq + r} = (a^k)^q . a^r

By hypothesis ab ≡ 1(mod n) and ak ≡ 1(mod n).

Hence, ar ≡ 1(mod n) where 0 ≤ r ≤ k

∴ r has to be equal to zero and otherwise the choice of k has the smallest positive integer such that ak ≡ 1(mod n) will be contradicted.

Hence, b = qk

⇒ k | b

⇒ k divides b

Hence, the correct answer is option 2)