 Hypersurface orthogonality implies symmetric extrinsic curvature

Unit normalization gives $n^{\nu }{\nabla }_{\mu }{n}_{\nu }=0$. Hence $K_{\mu \nu }=-P^{\alpha }{}_{\mu }{P}^{\beta }{}_{\nu }{\nabla }_{\alpha }{n}_{\beta }=-P^{\alpha }{}_{\mu }{\nabla }_{\alpha }{n}_{\nu }$. Set $F_{\mu \nu }={\nabla }_{\mu }{n}_{\nu }-{\nabla }_{\nu }{n}_{\mu }$. Frobenius theorem for a hypersurface normal gives $n_{\alpha }{F}_{\beta \gamma }+n_{\beta }{F}_{\gamma \alpha }+n_{\gamma }{F}_{\alpha \beta }=0$. Contracting with $n^{\alpha }{P}^{\beta }{}_{\mu }{P}^{\gamma }{}_{\nu }$ leaves $-P^{\beta }{}_{\mu }{P}^{\gamma }{}_{\nu }{F}_{\beta \gamma }=0$, precisely the vanishing antisymmetric part of the extrinsic curvature of a spatial hypersurface.