I apologize in advance for the slightly vague nature of this question.
If one adds a unary predicate $T$ to the language of arithmetic, one gets a sentence $L$ with the property that $L \leftrightarrow \neg T (\ulcorner L \urcorner)$ is a theorem of PA, by the usual fixed point construction.
Suppose however we are in the context of a 3-valued logic in which we may not assume that we have $T(x) \vee \neg T(x)$ for all $x$. I'm happy to assume that all the other relations / functions of arithmetic behave in a completely 2-valued way.
Do we still get a sentence $L$ for which $L \leftrightarrow \neg T (\ulcorner L \urcorner)$ is a theorem of PA in this 3-valued logic (however we flesh this out?)
In essence, the question is precisely how the Liar paradox is avoided by moving to 3-valued logic. Of course, it may well be the case that 3-valued logic avoids a contradiction in the presence of an $L$ for which the above biconditional is provable in virtue of the failure of excluded middle. That's great. But is it possible that the contradiction is blocked even prior to this in the sense that we don't even get a sentence $L$ for which the biconditional in question is provable in the first place?