Infinitesimal neighbourhoods and simultaneous normalization

Question

Let $B$ be a local, complete, integral $\mathbb{C}$-algebra of Krull dimension $1$ and $n:B \to \mathbb{C}[[t]]$ the normalization map. Given any local artinian $\mathbb{C}$-algebra $A$, we say that an infinitesimal deformation $B_A$ of $B$ admits simultaneous normalization if there exists a finite, injective morphism $$n_A:B_A \to A[[t]]$$ such that the restriction of $n_A$ to the special fiber is isomorphic to $n$. Does there exist an integer $N_0$, such that for all $N>N_0$, given any two infinitesimal deformations $B_A$ and $B'_A$ of $B$ which admits simultaneous normalizations $n_A:B_A \to A[[t]]$ and $n'_A:B'_A \to A[[t]]$, over the local artinian ring $A$, we have $B_A/K_A \cong B'_A/K'_A$ if and only if $B_A \cong B'_A$, where $$K_A:=\mbox{ker}(B_A \xrightarrow{n_A} A[[t]] \to A[[t]]/(t^N)) \mbox{ and } K'_A:=\mbox{ker}(B'_A \xrightarrow{n'_A} A[[t]] \to A[[t]]/(t^N))?$$ In other words, there exists an integer $N_0$, such that every infinitesimal deformation of $B$ which admits a simultaneous normalization is uniquely determined by its $N_0$-th order infinitesimal neighborhood, in the sense mentioned above.