Idempotents of a given rank form a closed set

Trace equals rank. Let $P^2=P$. Then $X^2-X=X(X-1)$ annihilates $P$, which has simple roots, so $P$ is diagonalisable with eigenvalues in $\{0,1\}$. If the eigenvalue $1$ has multiplicity $m$, then $\operatorname{rank}P=m$ (the image is the $1$-eigenspace) and $\operatorname{tr}P=m\cdot1+(n-m)\cdot0=m$. So $\operatorname{tr}P=\operatorname{rank}P$.

Description of $\mathcal{P}_k$. Therefore
$$\mathcal{P}_k=\{P:P^2=P\}\cap\{P:\operatorname{tr}P=k\}.$$
Indeed, "$\subseteq$" holds by the trace-equals-rank fact, and "$\supseteq$" because for an idempotent, $\operatorname{tr}P=k$ means $\operatorname{rank}P=k$.

Closed. The first set is the zero set of the polynomial map $P\mapsto P^2-P$, the second the preimage of $\{k\}$ under the continuous map $\operatorname{tr}$. Their intersection is closed.

Remark. The set $\{P^2=P\}$ has $n+1$ connected components $\mathcal{P}_0,\dots,\mathcal{P}_n$: since the trace is continuous and integer-valued on it, $\mathcal{P}_k$ are open in $\{P^2=P\}$ as well, and each is closed. Lesson. Integer-valued continuous functions (like the rank of an idempotent) separate components.