@Listing is correct.
The question tests to see if one recognizes that
= *O*(n) + *O*(1) + *O*(k')
= *O*(n) + *O*(1) + *O*(2^n)
<= M|n| + C|1| + N|k'| as x goes to infinity, and some constant M and N that satisfy:
(a) |work in part one| <= M|n|
(b) |definition of part two| <= C|1| (by assumption)
(c) |work in part three| <= N|k'|
as given or assumed.
<= M|n| + C|1| + N|2^n| since |k'| <= 1|2^n| as x goes to infinity (all k' are bounded above by 2^n by the definition of mod)
<= N|2^n| since | M|n| + C|1| + N|2^n| | <= 2N|2^n| as x goes to infinity
We note that if f(n,k) were to be O(n), then it would also be O(2^n); however, in this case, it is not O(n) since (as @Listing stated) k' can reach values much greater than n since k' takes on all values up to 2^n-1 (eg, when k=2^n-1, k'=2^n-1), and |2^n-1| > A|n| for all constants A as n goes to infinity.
See this page for more information on big O notation.