Assuming that you know all the terminologies of the question, here are the answers:
Let V are the number of VERTICES and E are number of EDGES.
A: Time complexity of Hamiltonian circuit is O( $n^{2}2^{n}$) where n is number of vertices. Hence, NPH.
The Hamilton circuit problem is polynomial time reducible to 3-SAT.
Now, since 3-SAT is NP complete therefore, Hamilton circuit problem also become np complete. Means NP hard.
B: Time complexity of 0/1 Knapsack is O(nW) where W is size of knapsack.
While the decision problem is NP-complete, the optimization problem is NP-hard, its resolution is at least as difficult as the decision problem, and there is no known polynomial algorithm which can tell, given a solution, whether it is optimal (which would mean that there is no solution with a larger V, thus solving the NP-complete decision problem).
Source: https://en.wikipedia.org/wiki/Knapsack_problem
D: Time complexity of Graph coloring is O$(V^{2}+E)$
This is NP-Complete. Reason : https://math.stackexchange.com/questions/125136/how-is-the-graph-coloring-problem-np-complete
C: Time Complexity of finding bi-connected components of a graph is O$(V+E)$. It is straightforward P time taking complexity.
Hence C is the answer.