The notation in the question prevents me from discussing a sequence of sets or applying the Rieger–Nishimura operations in several spaces at once, hence I will write $R(P,X,m)$ for what you denote $P_m$.

There is such a $P\subseteq\def\R{\mathbb R}\R^k$ for each $k\ge1$. First, intuitionistic propositional logic is complete with respect to the topological semantics when the topological space is fixed as $\R^k$ (this is true for every dense-in-itself separable metrizable space, see A. Tarski, Der Aussagenkalkül und die Topologie, Journal of Symbolic Logic 4 (1939), no. 1, pp. 26–27). This by itself implies that for each $m$, there exists $P$ such that $R(P,\R^k,m)\ne\R^k$. Better yet, we can fix sets $P_m\subseteq(m,m+1)^k$ such that $R(P_m,(m,m+1)^k,m)\ne(m,m+1)^k$. It is easy to check that $R(P,X,m)\cap U=R(P\cap U,U,m)$ for any open $U\subseteq X$, thus taking $P=\bigcup_mP_m$, we have $R(P,\R^k,m)\ne\R^k$ for all $m$.

The notation in the question prevents me from discussing a sequence of sets or applying the Rieger–Nishimura operations in several spaces at once, hence I will write $R(P,X,m)$ for what you denote $P_m$.

There is such a $P\subseteq\def\R{\mathbb R}\R^k$ for each $k\ge1$. First, intuitionistic propositional logic is complete with respect to the topological semantics when the topological space is fixed as $\R^k$ (this is true for every dense-in-itself separable metrizable space, see A. Tarski, Der Aussagenkalkül und die Topologie, Fundamenta Mathematicae 31 (1938), no. 1, pp. 103–134). This by itself implies that for each $m$, there exists $P$ such that $R(P,\R^k,m)\ne\R^k$. Better yet, we can fix sets $P_m\subseteq(m,m+1)^k$ such that $R(P_m,(m,m+1)^k,m)\ne(m,m+1)^k$. It is easy to check that $R(P,X,m)\cap U=R(P\cap U,U,m)$ for any open $U\subseteq X$, thus taking $P=\bigcup_mP_m$, we have $R(P,\R^k,m)\ne\R^k$ for all $m$.