A walk consists of an alternating sequence of vertices and edges consecutive elements of which are incident, that begins and ends with a vertex.
In simple words, Walk from vertex $w$ to $v$ is sequence of “adjacent edges”, starting from $w,$ ending at $v,$ possibly with repetition of vertices or edges or both. Length of a walk is the number of edges on it.
The Adjacency Matrix of a Graph :
The adjacency matrix $M$ of a undirected graph is defined by numbering the vertices, say from $1$ up to $n$, and then putting $M_{ij} = M_{ji} = 1 $ if there is an edge from $i$ to $j$, and $M_{ij} = 0$ otherwise.
We can do the same for a digraph: putting $M_{ij} = 1$ if there is an edge from $i$ to $j$, and $M_{ij} = 0$ otherwise.
Refer here to understand Adjacency Matrix of a Graph :
https://en.wikipedia.org/wiki/Adjacency_matrix
http://www.maths.nuigalway.ie/~rquinlan/linearalgebra/section1-1.pdf
Powers of the Adjacency Matrix :
Let $G$ be a graph, $n \in \mathbb{N}$, Let $M$ be the adjacency matrix of $G.$
The powers of the adjacency matrix counts things. In particular,
Entry $i, j$ in $M^n$ gives the number of walks from $i$ to $j$ of length $n.$
The proof is by induction argument. For example, the number of walks of length 2 is the number of vertices $k$ such that there is an edge from $i$ to $k$ and an edge from $k$ to $j$. And this is exactly the $i, j$ entry in $M^2$ , by the definition of matrix multiplication.
Also, NOTE that for any graph $G$ (directed or undirected), we can see that, every entry in adjacency matrix basically tells us that Number of walks of length $1$ between the corresponding vertices of that entry, So, if $M_{ij} $ is 1, it means there is walk of length $1$ from $i$ to $j,$ which is basically edge from $i$ to $j.$
Coming to the given question, for undirected connected graph $G,$ $M$ is its adjacency matrix. So, every non-zero entry $M_{ij}$ in $M$ tells us that there is a walk of length $1$ between vertices $i$ and $j$ in $G$.
Given $P = M^2,$ So, every non-zero entry $P_{ij}$ in $P$ tells us that there is a walk of length $2$ between vertices $i$ and $j$ in $G$.
( In particular, every entry $P_{ij}$ in $P$ tells us the number of walks of length $2$ between vertices $i$ and $j$ in $G$. )
By the definition of matrix $N$, $N_{ij} $ is $1$ iff either $M_{ij}$ is $1$ Or $P_{ij}$ is $1,$ So, we can say that $N_{ij} $ is $1$ iff there is a walk of length 1 or 2 or both between vertices $i,j$ in $G.$
$N$ is adjacency matrix of graph $G_2, $ So, in Graph $G_2,$ we have edge between $i,j$ iff there is walk of length 1 or 2 between $i,j$ in $G.$
So, conclusive point is that :
To get $G_2$ from $G,$ we need to keep the $G$ as it is, but also we need to add edge between vertex $w,u$ if there is walk of length 2 between $w,u.$ For undirected connected graph of at least two vertices, it will also bring self loops for every vertex.
So, diameter of $G_2$ will definitely become almost half of diameter of $G.$
If Diameter of $G$ is odd number, say $d =7,$ then Diameter of $G_2$ will become $\lceil d/2 \rceil = 4.$
If Diameter of $G$ is even number, say $d =6,$ then Diameter of $G_2$ will become $\lceil d/2 \rceil = 3.$
Option A is correct. But I think (Point me out if I am wrong) that $Diam(G_2) = \lceil Diam(G)/2 \rceil.$
http://www.maths.nuigalway.ie/~rquinlan/linearalgebra/section1-1.pdf
https://www.cusb.ac.in/images/cusb-files/2020/el/math/Lectures_on_Matrices_of_Graphs.pdf