= Solution
For an eigenvector $v$ with eigenvalue $\theta\in\{\lambda,\mu\}$,
$$
Bv=(A-\lambda I)(A-\mu I)v
=(\theta-\lambda)(\theta-\mu)v=0.
$$
By part (a), every vector orthogonal to $\mathbf e$ is a linear combination of such eigenvectors, so $B$ vanishes on $\mathbf e^\perp$. On the remaining eigenspace,
$$
B\mathbf e=(d-\lambda)(d-\mu)\mathbf e.
$$
The <orthogonal projection> onto $\operatorname{span}\{\mathbf e\}$ is $J/n$. It follows that
$$
\boxed{B=\frac{(d-\lambda)(d-\mu)}nJ}.
$$
Put $\kappa=(d-\lambda)(d-\mu)/n$. For $i\ne j$, the identity $B=\kappa J$ reads
$$
(A^2)_{ij}-(\lambda+\mu)A_{ij}=\kappa,
$$
and therefore
$$
(A^2)_{ij}=
\begin{cases}
\kappa+\lambda+\mu,&i\text{ and }j\text{ adjacent},\\
\kappa,&i\text{ and }j\text{ nonadjacent}.
\end{cases}
$$
Because $(A^2)_{ij}$ is the number of common neighbours, these two constants prove that $G$ is strongly regular. This is the converse half of the <three-eigenvalue characterization of a connected strongly regular graph>.
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