Let $X$ be a Banach space and $P : \mathcal{D}(P) \subset X \to X$ such that $P\mathcal{D}(P)\subset \mathcal{D}(P)$ and that $P^2x = Px,~\forall x \in \mathcal{D}(P).$
Does this imply that $P$ is bounded?
I really think that the answer is no. But could not handle an example so far. However, here is the intuition: suppose that $\{x_n\} \subset \mathcal{D}(P)$ converges to some $x\in \mathcal{D}(P)$. Then, $$P(P-I)x_n = 0,$$ and hence, for each $n$ there is $y_n \in \ker P$ satisfying $$Px_n = x_n + y_n.$$
Therefore, if $(y_n)$ does not converge, we are done.
Any hints?
Take $f:X\to\mathbb C$ linear and unbounded. Fix $x_0\in X$ with $f(x_0)=1$. Now define $Px=f(x)\,x_0$. Then $P$ is an unbounded projection of rank one.
You can make this natural if you don't require $P$ to be everywhere defined. For instance let $X=C[0,1]$ with the uniform norm, and let $\mathcal D(P)=C^1[0,1]$. Define $Pf$ to be the function $(Pf)t=f'(1)t$. Then $P$ is an unbounded projection.