= Solution
Use <metric compatibility> and differentiate the projector:
$$
\nabla_\alpha P^\epsilon{}_{\beta}
=(\nabla_\alpha n^\epsilon)n_\beta+n^\epsilon\nabla_\alpha n_\beta.
$$
Projecting the <derivative> and lower <tensor> index kills the first term, by part (a). The remaining term is the <extrinsic curvature>:
$$
\boxed{P^\alpha{}_{\mu}P^\beta{}_{\nu}\nabla_\alpha P^\epsilon{}_{\beta}
=n^\epsilon P^\alpha{}_{\mu}P^\beta{}_{\nu}\nabla_\alpha n_\beta
=K_{\mu\nu}n^\epsilon.}
$$
The second projection may be inserted in the definition of $K$, since differentiating $n^\beta n_\beta=-1$ gives $n^\beta\nabla_\alpha n_\beta=0$.
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