 Solution

Decompose each index with ${\delta }^{\rho }{}_{\mu }={\perp }^{\rho }{}_{\mu }-n^{\rho }{n}_{\mu }$. Since $n^{\nu }{n}_{\nu }=-1$, one has $n^{\nu }{\nabla }_{\mu }{n}_{\nu }=0$, so the second index of ${\nabla }_{\mu }{n}_{\nu }$ is spatial. Its spatial projection in the first index is $-K_{\mu \nu }$ by definition, while its normal projection isHence