C. For a problem to be NP-Complete, it must be NP-hard and it must also be in NP.
Ram's reduction shows that $\Pi$ is NP hard because it must be at least as hard as $3$-SAT which is a known NP-Complete problem. Here, $\Pi$ need not be NP-Complete.
Now, Shyam's reduction shows that $3$-SAT problem is at least as hard as $\Pi$ or equivalently $\Pi$ is not harder than $3$-SAT. Since $3$-SAT is in NP, $\Pi$ must also be in NP.
Thus, $\Pi$ is NP-hard and also in NP $\implies$ $\Pi$ is NP-Complete.