Blow up to get a morphism $\Pi: Bl_{P_0}\mathbf P^n \rightarrow \mathbf P^{n-1}$. Let $\widetilde{V}$ be the proper transform of $V$ in $Bl_{P_0}\mathbf P^n$. Then $\overline{\pi(V)}=\Pi(\widetilde{V})$.

Now we can write $\widetilde{V}=V \setminus \{P_0\} \ \cup \mathbf P(C_{P_O}V)$ where $C_{P_0}V$ is the tangent cone of $V$ at $P_0$.

So $\pi(V \setminus \{P_0\})$ (which in your notation is $\pi(V)$) contains $\Pi(\widetilde{V}) \setminus \Pi (\mathbf P(C_{P_O}V))$.

As noted above, $\Pi(\widetilde{V})$ equals $\overline{\pi(V)}$. Moreover, $\mathbf P(C_{P_O}V))$ is a closed subset of the exceptional divisor $E$, and $\Pi_{|E} \colon E \rightarrow \mathbf P^{n-1}$ is an isomorphism.

So we get that $\pi(V)$ (in your notation) contains $\overline{\pi(V)} \setminus W$ where $W \subset \mathbf P^{n-1}$ is a closed subset isomorphic to the projectivisation of the tangent cone of $V$ at $P_0$.

The closed set $W$ has dimension $\operatorname{dim}(V)-1$. On the other hand, $\pi(V)$ has the same dimension as $V$ unless $V$ is a cone whose vertex contains $P_0$, but in that case $\pi(V)$ is a closed set .

As for degree, the degree of $\mathbf P(C_{P_O}V))$ as a *subscheme* of $E$ equals the multiplicity of $V$ at $P_0$, hence is bounded above by $\operatorname{deg}(V)$. Since $W$ is (isomorphic to) the underlying closed subset of this scheme, its degree is not greater than that of the scheme. So we have $\operatorname{deg}(W) \leq \operatorname{deg}(V)$ as required.