Let $G$ be a finite, simple graph without loop. Assume that, for some $r ≥ 2$, it is possible to find a set of $r$ vertices all having the same neighbors. To show that $0$ is an eigenvalue of $A$, with multiplicity at least $r − 1$.
Require some hints for the problem.
Order our vertices $v_1, \ldots, v_r, v_{r+1}, \ldots, v_n$ where $N(v_i)=N(v_j)$ for all $1 \leq i \leq j \leq r$ i.e. the vertices $v_1, \ldots, v_r$ have the same neighbours. If for some $i,j \leq r$ we have $v_i v_j \in E(G)$ then as $v_i \in N(v_j)$ we must have $v_i \in N(v_i)$ and $v_i v_i \in E(G)$ which contradicts that $G$ has no loops.
We now note that for the adjacency matrix $A$ with rows and columns ordered as the vertices are we have that the top left $r \times r$ submatrix of $A$ is $0_r$. We now note that $$A=\begin{bmatrix}0_r&B\\B^T&C\end{bmatrix}$$ where $B$ is a $r \times (n-r)$ matrix where every row is some vector $x:=(x_{r+1}, \ldots, x_n)$ and $D$ is a $(n-r) \times (n-r)$ symmetric matrix. With this we note that $$\begin{bmatrix}0_r&B\\B^T&C\end{bmatrix} \begin{bmatrix}y_1\\y_2\end{bmatrix} = \begin{bmatrix}B y_1\\B^Ty_1 + Cy_2\end{bmatrix}$$ for any $y_1 \in \mathbb{R}^r$, $y_2 \in \mathbb{R}^{n-r}$. It can be seen that $$By_1 = \begin{bmatrix}xy_1\\ \vdots \\ xy_1\end{bmatrix},$$ thus the image of $A$ has dimension at most $n-r+1$ and thus $\ker A$ has dimension at least $r-1$. This is equivalent to $0$ being an eigenvalue of $A$ with multiplicity at least $r-1$.