= Solution
If $p=m\times n$, then $n^Tp=0$, and hence $Bp=p$. Also $Bm=m$ because $m\perp n$. Thus $p$ and $m$ are linearly independent eigenvectors with eigenvalue one.
For every vector $v$,
$$
(B-I)v=m(n^Tv),
$$
so $B-I$ is nonzero but has square zero:
$$
(B-I)^2=mn^Tmn^T=0.
$$
Consequently all three eigenvalues are one. The eigenspace is
$$
\ker(B-I)=\{v:n^Tv=0\}=n^\perp,
$$
which has dimension two. It cannot supply three linearly independent eigenvectors, so
$$
\boxed{B\text{ has eigenvalues }1,1,1\text{ and is not diagonalizable}.}
$$
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