Let $G$ be a connected reductive group of an arbitrary field $k$, $P$ a parabolic $k$-subgroup, and $U = \mathscr{R}_u(P)$ the unipotent radical of $P$ (so $U$ is what is unfortunately traditionally denoted as $N$). We have that $U(k)$-conjugation is simply transitive on the set of all Levi $k$-subgroups of $P$, so to answer your question it suffices to treat a single Levi $k$-subgroup of $P$ (in the case of a non-archimedean field).

By the Borel-Tits structure theory that is valid over all fields (and explained in Borel's textbook on algebraic groups), if we choose a maximal split $k$-torus $S$ in $P$ then it is also maximal as such in $G$, and there is a minimal parabolic $k$-subgroup $B$ of $G$ contained in $P$ and containing $S$, with $\Phi(B,S)$ a positive system of roots in the relative root system ${}_k\Phi = \Phi(G,S)$.

Letting $\Delta$ be the base of $\Phi(B,S)$, it is a basis of ${\rm{X}}(S)_{\mathbf{Q}}$. Let $\{a^{\ast}\}_{a \in \Delta}$ be the dual basis of ${\rm{X}}_{\ast}(S)_{\mathbf{Q}}$. Under the usual indexing of parabolic $k$-subgroups of $G$ containing $B$ (so-called "standard" parabolics) by subsets of $\Delta$, we have $P = P_I$ for a unique subset $I \subset \Delta$ (where $P_{\emptyset} = B$ and $P_{\Delta} = G$). Explicitly, we have the dynamic description $P_I = P_G(\lambda_I)$ where $$\lambda_I = N \cdot \sum_{a \not\in I} a^{\ast}$$ for a sufficiently divisible integer $N > 0$ such that $\lambda_I$ lies in the lattice ${\rm{X}}_{\ast}(S)$. (The choice of $N$ doesn't matter, since $P_G(\lambda) = P_G(n \lambda)$ for any $n > 0$ and $\lambda \in {\rm{X}}_{\ast}(S)$).

We have the Levi $k$-subgroup $M = Z_G(\lambda_I)$ of $P$ and also $U = U_G(\lambda_I)$, so in particular $\lambda_I(t) \in S(k)$ for any $t \in k^{\times}$. The $k$-group $U$ is directly spanned in any order by the root groups $U_a$ for non-multipliable $a \in {}_k\Phi$ such that $\langle a, \lambda_I \rangle > 0$. (Of course, ${}_k\Phi$ might be non-reduced, and $\dim U_a$ might be very large, and $U_a$ might be non-commutative if $a$ is divisible.)

So in view of the explicit description of $\lambda_I(t)$-conjugation on $U_a$ (namely, a vector group with action by $t^{\langle a, \lambda_I\rangle}$ when $a$ isn't divisible, and an extension of such by another using $2a$ when $a$ is divisible), we see that any $t \in k^{\times}$ with $|t| < 1$ yields $s := \lambda_I(t) \in S(k)$ that does the job (as $\langle a, \lambda_I \rangle > 0$ precisely for those $a$ occurring as $S$-weights on ${\rm{Lie}}(U)$).

Let $G$ be a connected reductive group of an arbitrary field $k$, $P$ a parabolic $k$-subgroup, and $U = \mathscr{R}_u(P)$ the unipotent radical of $P$ (so $U$ is what is unfortunately traditionally denoted as $N$).

For any Levi $k$-subgroup $M$ of $P$ there is a 1-parameter $k$-subgroup $\lambda:{\rm{GL}}_1 \rightarrow G$ such that $P = P_G(\lambda)$, $M = Z_G(\lambda)$, and $U = U_G(\lambda)$. In particular, $\lambda$ factors through the maximal split central $k$-torus $S_M$ in $M$, so $\lambda(t) \in S_M(k)$ for any $t \in k^{\times}$. By the Borel-Tits structure theory, if $S$ is a maximal split $k$-torus in $M$ containing $S_M$ then it is also maximal as such in $G$ and the set $\Phi(G,S)$ of nontrivial $S$-weights on ${\rm{Lie}}(G)$ is a root system (spanning ${\rm{X}}(S)_{\mathbf{Q}}$), possibly non-reduced.

The $k$-group $U$ is directly spanned in any order by the root groups $U_a$ for non-multipliable $a \in \Phi(G,S)$ such that $\langle a, \lambda \rangle > 0$. (Of course, $\dim U_a$ might be very large, and $U_a$ might be non-commutative if $a$ is divisible.)

If $a$ is not divisible in $\Phi(G,S)$ then $U_a$ is a vector group admitting a linear structure relative to which the effect of $\lambda(t)$-conjugation on $U_a$ is scaling by $t^{\langle a, \lambda\rangle}$, and is an extension of such by another using $2a$ when $a$ is divisible. In this way we see that any $t \in k^{\times}$ with $|t| < 1$ yields $s := \lambda(t) \in S_M(k)$ that does the job (as $\langle a, \lambda \rangle > 0$ precisely for those $a$ occurring as $S$-weights on ${\rm{Lie}}(U)$).

Let $G$ be a connected reductive group of an arbitrary field $k$, $P$ a parabolic $k$-subgroup, and $U = \mathscr{R}_u(P)$ the unipotent radical of $P$ (so $U$ is what is unfortunately traditionally denoted as $N$). We have that $U(k)$-conjugation is simply transitive on the set of all Levi $k$-subgroups of $P$, so to answer your question it suffices to treat a single Levi $k$-subgroup of $P$ (in the case of a non-archimedean field).

By the Borel-Tits structure theory that is valid over all fields (and explained in Borel's textbook on algebraic groups), if we choose a split maximal $k$-torus $S$ in $P$ then it is also maximal as such in $G$, and there is a minimal parabolic $k$-subgroup $B$ of $G$ contained in $P$ and containing $S$, with $\Phi(B,S)$ a positive system of roots in the relative root system ${}_k\Phi = \Phi(G,S)$.

Letting $\Delta$ be the base of $\Phi(B,S)$, it is a basis of ${\rm{X}}(S)_{\mathbf{Q}}$. Let $\{a^{\ast}\}_{a \in \Delta}$ be the dual basis of ${\rm{X}}_{\ast}(S)_{\mathbf{Q}}$. Under the usual indexing of parabolic $k$-subgroups of $G$ containing $B$ (so-called "standard" parabolics) by subsets of $\Delta$, we have $P = P_I$ for a unique subset $I \subset \Delta$ (where $P_{\emptyset} = B$ and $P_{\Delta} = G$). Explicitly, we have the dynamic description $P_I = P_G(\lambda_I)$ where $$\lambda_I = N \cdot \sum_{a \not\in I} a^{\ast}$$ for a sufficiently divisible integer $N > 0$ such that $\lambda_I$ lies in the lattice ${\rm{X}}_{\ast}(S)$. (The choice of $N$ doesn't matter, since $P_G(\lambda) = P_G(n \lambda)$ for any $n > 0$ and $\lambda \in {\rm{X}}_{\ast}(S)$).

We have the Levi $k$-subgroup $M = Z_G(\lambda_I)$ of $P$ and also $U = U_G(\lambda_I)$, so in particular $\lambda_I(t) \in S(k)$ for any $t \in k^{\times}$. The $k$-group $U$ is directly spanned in any order by the root groups $U_a$ for non-multipliable $a \in {}_k\Phi$ such that $\langle a, \lambda_I \rangle > 0$. (Of course, ${}_k\Phi$ might be non-reduced, and $\dim U_a$ might be very large, and $U_a$ might be non-commutative if $a$ is divisible.)

So in view of the explicit description of $\lambda_I(t)$-conjugation on $U_a$ (namely, a vector group with action by $t^{\langle a, \lambda_I\rangle}$ when $a$ isn't divisible, and an extension of such by another using $2a$ when $a$ is divisible), we see that any $t \in k^{\times}$ with $|t| < 1$ yields $s := \lambda_I(t) \in S(k)$ that does the job (as $\langle a, \lambda_I \rangle > 0$ precisely for those $a$ occurring as $S$-weights on ${\rm{Lie}}(U)$).

Let $G$ be a connected reductive group of an arbitrary field $k$, $P$ a parabolic $k$-subgroup, and $U = \mathscr{R}_u(P)$ the unipotent radical of $P$ (so $U$ is what is unfortunately traditionally denoted as $N$). We have that $U(k)$-conjugation is simply transitive on the set of all Levi $k$-subgroups of $P$, so to answer your question it suffices to treat a single Levi $k$-subgroup of $P$ (in the case of a non-archimedean field).

By the Borel-Tits structure theory that is valid over all fields (and explained in Borel's textbook on algebraic groups), if we choose a maximal split $k$-torus $S$ in $P$ then it is also maximal as such in $G$, and there is a minimal parabolic $k$-subgroup $B$ of $G$ contained in $P$ and containing $S$, with $\Phi(B,S)$ a positive system of roots in the relative root system ${}_k\Phi = \Phi(G,S)$.

Letting $\Delta$ be the base of $\Phi(B,S)$, it is a basis of ${\rm{X}}(S)_{\mathbf{Q}}$. Let $\{a^{\ast}\}_{a \in \Delta}$ be the dual basis of ${\rm{X}}_{\ast}(S)_{\mathbf{Q}}$. Under the usual indexing of parabolic $k$-subgroups of $G$ containing $B$ (so-called "standard" parabolics) by subsets of $\Delta$, we have $P = P_I$ for a unique subset $I \subset \Delta$ (where $P_{\emptyset} = B$ and $P_{\Delta} = G$). Explicitly, we have the dynamic description $P_I = P_G(\lambda_I)$ where $$\lambda_I = N \cdot \sum_{a \not\in I} a^{\ast}$$ for a sufficiently divisible integer $N > 0$ such that $\lambda_I$ lies in the lattice ${\rm{X}}_{\ast}(S)$. (The choice of $N$ doesn't matter, since $P_G(\lambda) = P_G(n \lambda)$ for any $n > 0$ and $\lambda \in {\rm{X}}_{\ast}(S)$).

We have the Levi $k$-subgroup $M = Z_G(\lambda_I)$ of $P$ and also $U = U_G(\lambda_I)$, so in particular $\lambda_I(t) \in S(k)$ for any $t \in k^{\times}$. The $k$-group $U$ is directly spanned in any order by the root groups $U_a$ for non-multipliable $a \in {}_k\Phi$ such that $\langle a, \lambda_I \rangle > 0$. (Of course, ${}_k\Phi$ might be non-reduced, and $\dim U_a$ might be very large, and $U_a$ might be non-commutative if $a$ is divisible.)

So in view of the explicit description of $\lambda_I(t)$-conjugation on $U_a$ (namely, a vector group with action by $t^{\langle a, \lambda_I\rangle}$ when $a$ isn't divisible, and an extension of such by another using $2a$ when $a$ is divisible), we see that any $t \in k^{\times}$ with $|t| < 1$ yields $s := \lambda_I(t) \in S(k)$ that does the job (as $\langle a, \lambda_I \rangle > 0$ precisely for those $a$ occurring as $S$-weights on ${\rm{Lie}}(U)$).