$\newcommand\1{\mathbf1}\newcommand{\R}{\mathbb R}\newcommand{\la}{\lambda}$Here is yet another, "multivariate calculus" way to treat your integral, with the same result as in my other two answers on this page.
Let $$c:=N^2r\in(0,\infty),$$ $$\Pi_x:=\{y\in\R^k\colon \1_k\cdot y=x\},$$ where $\cdot$ denotes the dot product and $\1_k:=(1,\dots,1)\in\R^k$.
The integral in question is \begin{equation*} I_x:=e^{cx^2/k}J_x,\quad\text{where}\quad J_x:=\int_{\Pi_x}\mu_x(dy)e^{-c|y|_k^2}, \tag{1} \end{equation*} $|y|_k$ is the Euclidean norm of $y\in\R^k$, and $\mu_x(dy)$ is the surface area element on the plane $\Pi_x$. This plane is the graph of the function $f_x\colon\R^{k-1}\to\R$ given by the formula $f_x(y):=x-y_1-\cdots-y_{k-1}$ for $y=(y_1,\dots,y_{k-1})\in\R^{k-1}$. So (cf. the case $k=3$), \begin{align*} J_x&=\int_{\R^{k-1}}dy_1\cdots dy_{k-1}\,\sqrt{1+\sum_{j=1}^{k-1}\Big(\frac{\partial f_x}{\partial y_j}\Big)^2} \\ & \times \exp\Big\{-c(x-y_1-\cdots-y_{k-1})^2-c\sum_{j=1}^{k-1}y_j^2\Big\} \\ &=\sqrt k\,\int_{\R^{k-1}}dy\, \exp\big\{-c(x-y\cdot\1_{k-1})^2-c|y|_{k-1}^2\big\} \\ &=\sqrt k\,\int_{\R^{k-1}}dz\, \exp\big\{-c(x-z_1\sqrt{k-1})^2-c|z|_{k-1}^2\big\}; \end{align*} for the last displayed equality, we use any substitution of the form $y=Qz$, where $Q$ is any orthogonal $(k-1)\times(k-1)$ matrix whose first column is $\1_{k-1}/\sqrt{k-1}$ and $z=(z_1,\dots,z_{k-1})$, so that $z_1=y\cdot\1_{k-1}/\sqrt{k-1}$.
So, \begin{align*} J_x&=\sqrt k\,\int_\R dz_1\, \exp\big\{-c(x-z_1\sqrt{k-1})^2-cz_1^2\big\} \\ &\times\int_{\R^{k-2}}dz_2\cdots dz_{k-1}\, \exp\Big\{-c\sum_{j=2}^{k-1}z_j^2\Big\} \\ &=\sqrt k\ \;\frac{e^{-cx^2/k}(\pi/c)^{1/2}}{\sqrt k}\ \;(\pi/c)^{(k-2)/2} \\ &=e^{-cx^2/k}\;(\pi/c)^{(k-1)/2}. \end{align*}
Thus, by (1), the integral in question is $$I_x=(\pi/c)^{(k-1)/2}=\Big(\frac\pi{N^2r}\Big)^{(k-1)/2},$$ which is what was obtained a bit differently in my other two answers.