= Solution
An <eigenvalue> can be complex even though $P$ is real, so use the complex <Hermitian inner product>. For a nonzero complex <eigenvector> $\mathbf v$, the real <orthogonal matrix> also satisfies $P^*P=P^TP=I$. Hence
$$
\|\mathbf v\|^2=\|P\mathbf v\|^2=|\lambda|^2\|\mathbf v\|^2,
$$
and \b[$\boxed{|\lambda|=1}$]. Taking complex conjugates in $P\mathbf v=\lambda\mathbf v$ gives $P\overline{\mathbf v}=\overline\lambda\,\overline{\mathbf v}$ because $P$ is real. The conjugate vector is nonzero, so \b[$\boxed{\lambda^*=\overline\lambda\text{ is also an eigenvalue}}$].
For the unheaded geometrical conclusion, the third <eigenvalue> of $Q$ must be real: nonreal <eigenvalues> of a real matrix occur in conjugate pairs. Its modulus is one, so it is $1$ or $-1$. Exact <algebraic multiplicity> two for $1$ excludes a third $1$, leaving $-1$. By the supplied diagonalisability fact, the $1$-<eigenspace> is a two-dimensional <plane> and the $-1$-<eigenspace> is a line. They are <orthogonal>: preservation of the <inner product> between a fixed vector and a negated vector gives $\mathbf u\cdot\mathbf v=-\mathbf u\cdot\mathbf v$. Therefore \b[$Q$ is the <orthogonal reflection> in its fixed <plane>]. It fixes the two tangential directions and reverses the normal direction; with unit normal $\mathbf n$, $Q=I-2\mathbf n\mathbf n^T$.
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