= Solution
Write $q_{\alpha\beta}=g_{\alpha\beta}+n_\alpha n_\beta$ for the <induced metric>. The <spatial covariant derivative> projects its derivative index and both tensor indices:
$$
D_\gamma q_{\alpha\beta}=P^\rho{}_\gamma P^\mu{}_\alpha P^\nu{}_\beta\nabla_\rho(g_{\mu\nu}+n_\mu n_\nu).
$$
Four-dimensional <metric compatibility> removes the derivative of $g$. Each remaining term contains an undifferentiated normal on a projected index and hence vanishes. Therefore
$$
\boxed{D_\gamma q_{\alpha\beta}=0.}
$$
This establishes <metric compatibility of the spatial covariant derivative> directly from its projected definition.
Differentiating the normal's normalization gives
$$
0=\nabla_\nu(n^\mu n_\mu)=2n^\mu\nabla_\nu n_\mu,
$$
so \b[$n^\mu\nabla_\nu n_\mu=0$]. Consequently the second projection in the <extrinsic curvature> definition is redundant:
$$
K_{\alpha\beta}=P^\mu{}_\alpha(\delta^\nu{}_\beta+n^\nu n_\beta)\nabla_\mu n_\nu
=P^\mu{}_\alpha\nabla_\mu n_\beta.
$$
Finally differentiate $P^\lambda{}_\nu=\delta^\lambda{}_\nu+n^\lambda n_\nu$ and project:
$$
P^\mu{}_\alpha P^\nu{}_\beta\nabla_\mu P^\lambda{}_\nu
=P^\mu{}_\alpha P^\nu{}_\beta[(\nabla_\mu n^\lambda)n_\nu+n^\lambda\nabla_\mu n_\nu]
=\boxed{K_{\alpha\beta}n^\lambda.}
$$
The first term vanishes by $P^\nu{}_\beta n_\nu=0$, and the second is precisely the defining curvature. The positive sign follows from the specified positive-expansion convention, rather than from an independent projector convention.
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