 Solution

Write $\epsilon =n_a{n}^a\in \{-1,+1\}$. The non-null hypersurface projection is $P^a{}_b={\delta }^a{}_b-\epsilon n^a{n}_b$, reproducing both signs in the question. It annihilates the normal and is the identity on tangent vectors to the hypersurface.

Represent $V$ by the tangent to a curve $\lambda$ through $p$. The curve $\varphi \circ \lambda$ lies in $\Sigma$, so its tangent $d{\varphi }_{p}V$ is tangent to $\Sigma$. Orthogonality therefore gives $n_a({\varphi }_{\ast }V{)}^a=0$, andThis uses the stated smooth hypersurface assumption; the dimension relation alone would not make the image of an arbitrary smooth map a regular hypersurface.