$, then $A\in{\boldsymbol{\text{S}{{\mathcal{N}}}}}_\kappa$.  Proof: Fix an enumeration of $B$, $B=\{b_\alpha\colon \alpha<\kappa\}$. Let $I=\{\xi_\alpha\colon \alpha<\kappa\}\in[\kappa]^\kappa$, and let $f\colon \kappa\times\{0,0\}\to \kappa$ be a bijection. 