Consider the following primal problem: $$ (1) \quad \max_{y} c^\top y ,\\ \quad \quad \quad \text{s.t. } By \leq a $$ The dual of (1) is $$ (2) \quad \min_{x\geq 0} a^\top x,\\ \quad \quad \quad \text{s.t. }B^\top x=c $$
Consider the Lagrangian of (2): $$ L(x,\mu,v)=a^\top x-\nu^\top x+\mu^\top(B^\top x-c) $$
Consider the sets of $x$ and $\mu$ satisfying the KKT conditions of (2). Let me call these sets $\mathcal{X}$ and $\mathcal{M}$ respectively. I want to perturb (2) so as to ensure that $\mathcal{X}$ and $\mathcal{M}$ are singleton. Following this suggestion here, I replace (2) with
$$ (3) \quad \min_{x\geq 0} a^\top x+ \epsilon ||x||_2,\\ \quad \quad \quad \text{s.t. }B^\top x=c $$ with $\epsilon>0$. If $\epsilon$ is small enough, then (3) $\approx $ (2).
Questions:
Are $\mathcal{X}$ and $\mathcal{M}$ of (3) singleton, as desired?
What is the primal that has (3) as dual?
You could consider $$ \min_{x\geq 0} a^\top x+ \epsilon_1 ||x||_2,\\ \quad \text{s.t. } ||B^\top x-c||_2 \leq \epsilon_2 $$
The Lagrange dual is \begin{align} & \max_{\lambda\geq 0,\nu \geq 0} \min_{x} a^\top x+ \epsilon_1 ||x||_2 + \lambda(||B^\top x-c||_2 - \epsilon_2) - \nu^Tx \\ = \; & \max_{\lambda\geq 0,\nu \geq 0} \min_{x,z} \max_{y} a^\top x+ \epsilon_1 ||x||_2 + \lambda(||z||_2 - \epsilon_2) - \nu^Tx + y^T(z+c-B^\top x) \\ = \; & \max_{y,\lambda\geq 0,\nu \geq 0} \min_{x,z} (a-By-\nu)^T x+ \epsilon_1 ||x||_2 + y^Tz + \lambda||z||_2 + c^Ty - \lambda\epsilon_2 \\ = \; & \max_{y,\lambda\geq 0,\nu \geq 0} \left\{ c^Ty - \lambda\epsilon_2 : ||By-a+\nu||\leq\epsilon_1, ||y||_2 \leq \lambda \right\} \\ \end{align} To express the dual we needed an extra variable $y$ which is not necessarily unique, but the dual variable $\lambda$ is unique because the corresponding constraint in the primal ($||B^\top x-c||_2 \leq \epsilon_2$) satisfies the LICQ. I am therefore not sure if this is helpful, because you still have the 'same' $y$ you started with.
Either way, by taking $\epsilon_2=0$ you obtain the dual to (3): $$\max_{y,\nu \geq 0} \left\{ c^Ty : ||By-a+\nu||\leq\epsilon \right\}.$$