Lasso - constraint form equivalent to penalty form

Here is one direction.

(1) The constrained problem is of the form \begin{array}{ll} \text{Find} & x \\ \text{To minimize} & f(x) \\ \text{such that} & g(x) \leqslant t \\ & \llap{-} g(x) \leqslant t. \end{array} Its Lagrangian is $$ L(x, \mu_1, \mu_2) = f(x) + \mu_1' ( g(x) - t ) + \mu_2' ( - g(x) - t ) $$ and the KKT conditions are \begin{align*} \nabla f + \mu_1' \nabla g - \mu_2' \nabla g &= 0 \\ \mu_1, \mu_2 &\geqslant 0 \\ \mu_1' ( g(x) - t ) &= 0 \\ \mu_2' ( - g(x) - t ) &= 0 . \end{align*}

(2) The penalized problem is just the minimization of $f(x) + \lambda' g(x)$. It is unconstrained, and the first order condition is $$ \nabla f + \lambda ' \nabla g = 0. $$

Given a solution of the constrained problem, the penalized problem with $\lambda = \mu_1 - \mu_2$ has the same solution. (For a complete proof, you also need to check that, in your situation, the KKT conditions and the first order condition are necessary and sufficient conditions.)