Without further requirements on $\tau$, this is trivial. Let $T$ be any first-order theory and let $\top$ be any tautology. Define $\tau(\phi)= \top$. 

If you want $\tau$ to be injective, then let $T$ be any first-order theory in the language of $TA$ and define $\tau(\phi)=\top\vee\phi$.

If you also want equivalence, then take $T$ to be any first-order theory in the language of $TA$ (such that $T\subseteq TA$) and define $\tau$ as follows: 

-$\tau(\phi)=\phi$ if $\phi$ is not a theorem of $TA$

-$\tau(\phi)=\top\vee\phi$ if $\phi$ is a theorem of $TA$

This $\tau$ is injective and gives you the equivalence: 
$\forall \phi(TA\vdash\phi\leftrightarrow T\vdash\tau(\phi))$.

If you want $\tau$ to be an interpretation in the usual sense, as the title of the question suggests, and arithmetical, then this is not possible as Fedor Pakhomov has explained in two comments (intepretations preserve negation). 

So, you must specify what do you mean by a suitable $\tau$, as Monroe Eskew has already asked in a comment.