= Solution
Differentiate the normalization of the timelike <unit normal> with the metric-compatible <covariant derivative>:
$$
0=\nabla_\nu(n^\mu n_\mu)=2n^\mu\nabla_\nu n_\mu.
$$
Thus the <derivative> of the normal is already orthogonal to the normal in its second index. Expanding the second <spatial projection tensor> in the definition of the <extrinsic curvature of a spatial hypersurface> gives
$$
K_{\mu\nu}=-P^\alpha{}_\mu(\delta^\beta{}_\nu+n^\beta n_\nu)\nabla_\alpha n_\beta=-P^\alpha{}_\mu\nabla_\alpha n_\nu.
$$
To prove symmetry using <Frobenius theorem>, set $F_{\alpha\beta}=\nabla_\alpha n_\beta-\nabla_\beta n_\alpha$. <Hypersurface orthogonality> is equivalent to
$$
n_\alpha F_{\beta\lambda}+n_\beta F_{\lambda\alpha}+n_\lambda F_{\alpha\beta}=0.
$$
Contract this with $n^\alpha P^\beta{}_\mu P^\lambda{}_\nu$. The last two terms vanish because a projected normal vanishes, while $n^\alpha n_\alpha=-1$ in the first term. Hence
$$
P^\alpha{}_\mu P^\beta{}_\nu F_{\alpha\beta}=0.
$$
The antisymmetric part of the doubly projected <covariant derivative> of $n$ is therefore zero. This proves that <hypersurface orthogonality implies symmetric extrinsic curvature>:
$$
\boxed{n^\mu\nabla_\nu n_\mu=0,\quad K_{\mu\nu}=-P^\alpha{}_\mu\nabla_\alpha n_\nu,\quad K_{\mu\nu}=K_{\nu\mu}.}
$$
No geodesic assumption for the normal congruence is needed; its <normal acceleration> may be nonzero.
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