First we’ll understand the property $P$ that the question mentions –
Let $G = (V,E)$ be any undirected graph on $n$ nodes, the graph $G$ has property $P$ that every subset $S$ of $V$ such that $S$ having at most $\frac{n}{2}$ nodes then the closed neighbourhood of $S, N[S]$ has at least $1.5*|S|$ nodes.
That is, $(|S| = x) \land (x \le \frac{n}{2}) \implies |N[S]| = 1.5*x \implies$ the (open) neighbourhood of $S, |N(S)| = |S’| = 0.5*x$.
(A).
Consider a complete graph $K_n$ on $n$ nodes, in this graph every node has $(n-1)$ neighbours.
Closed neighbourhood of any subset $S$ of $V$ will have $n$ nodes i.e. $|N[S]| = n, \forall S \subset V$.
Therefore, $K_n$ has property $P$ and we know longest path in $K_n$ has length $n-1$.
Therefore, then length of longest path in $G$ is $O(n)$.
(B).
Lets assume, in a graph $G$ having property $P$ the length of the longest path is $O(\sqrt n)$.
That is, the length of the longest path is $k * \sqrt n$, where $k$ is some positive constant.
Now, we want every node of our graph $G$ to have maximum possible neighbours. The only way to do this is to create $\frac{\sqrt n}{k}$ connected components, each of order $k * \sqrt n$ such that each connected component is a complete subgraph.
Now, select $c$ components such that $c*k* \sqrt n \le \frac{n}{2}$, i.e. the total number of nodes in $c$ selected components is at most $\frac{n}{2}$.
Now, as per property $P$ the closed neighbourhood of this selection of nodes must have at least $1.5*c*k* \sqrt n$ nodes.
But, since any node in the $c$ components is not connected to any other node outside its own component, the closed neighbourhood of our selection will have only $c*k* \sqrt n$ nodes.
This is a contradiction.
Therefore, any graph $G$ having property $P$ cannot have largest path of length $O(\sqrt n)$.
Answer :- E.