What undecidable language $B$ is reducible to its complement?
Let $H$ be the language of all Turing machines that halt on empty input.
Clearly $H$ is undecidable.
Let $L = \{ (1,T) : T \in H \} \cup \{ (0,T) : T \not\in H \}$.
Clearly $L$ is undecidable. If $L$ were decidable, then a Turing machine $M$  for $L$ would also imply the existence of a Turing machine $M'$ that decides $H$. $M'$ with input $T$ simply simulates $M$ with input $(1,T)$.
Moreover, for a Turing machine $T$ and $x \in \{0,1\}$ we have:
$$
(x, T) \in L \iff (1-x, T) \not\in L \iff (1-x, T) \in \overline{L}.
$$
This, combined with the fact that we can decide whether a given word encodes a valid Turing machine, shows that $L$ is reducible to $\overline{L}$.