So the correct answer is A.

Let's consider each of the two graphs In turn.

Let's start with a star graph on n vertices.

So this has one vertex in the centre and n-1 spokes.

The claim is it has a vertex cover of size one, obviously the minimum possible.

That vertex cover comprises just that center vertex.

Why is it a vertex cover?

Well every single edge is a spoke and one of its endpoints is the center vertex.

So you just pick the one vertex and you hit all of the n-1 edges of the graph.

So how about a clique on n vertices?

So a graph in which every single one of the n choose two edges is present.

Why is n minus one vertices necessary and sufficient for a vertex cover?

To see why it's necessary,

consider any set that excludes at least two vertices of the graph.

Because it's a clique there's an edge between those two excluded vertices and

that's an edge such that neither of it's end points is in the set.

So that's why the set is not a vertex cover.

If it has the most N-2 vertices in it.

On the other hand if it only excludes one vertex

then there cannot be an edge with both of it's end points missing from the set.

because there's only one vertex in total missing for the set.

So that's why any subset of n- 1 vertices will be a vertex cover of a clique,

n- 1 is sufficient.