Let $H^1(\Omega)=\{v\in L^2(\Omega):\nabla v\in L^2(\Omega)^2\}$, where $L^2(\Omega)$ is the usual space of square-integrable functions.
There exists an operator $$P:H^1(\Omega)\longrightarrow P_k(\Omega)$$ such that $\|Pv\|{\color{red}\geq} C\|v\|$, with $C>0$ a constant independent of $v$? (note the red sign).
Here, $P_k(\Omega)$ is the set of polynomials defined on $\Omega$ less than $k\geq 0$, and the norm is:
$$\|v\|^2=\|v\|_0^2+\|\nabla v\|_0^2$$
for all $v\in H^1(\Omega)$, where $\|\cdot \|_0$ is the norm induced by the $L^2$ inner product:
$$\|v\|_0^2=\displaystyle\int_\Omega v^2\,dx$$
Note that $P$ is not an projector and then not requires to hold any projector propertie.
Let $p \in P_k(\Omega)$ be fixed with $\|p\| = 1$. Then, you can define $P(v) := \|v\| \, p$. This yields $\|P v \| = \|v\| \, \|p\| = \|v\|$. That is, your desired inequality holds with $C = 1$ (and is even an equality).