Is every degree above ${\bf 0''}$ PA over something close to itself?

The low basis theorem says that there are PA degrees which are low - that is, which satisfy $${\bf a'}={\bf 0'}$$. Appropriately relativized, given a degree $${\bf a}$$ there is a degree $${\bf b}$$ "not much bigger than" $${\bf a}$$ which is PA over $${\bf a}$$. A natural question at this point is whether the "dual" holds: if $${\bf b}$$ is PA in the first place, must $${\bf b}$$ be PA over some $${\bf a}$$ which is "close to" $${\bf a}$$? One natural way to phrase this question would be: if $${\bf b}$$ is PA, then must $${\bf b}$$ be PA over some $${\bf a}$$ with $${\bf a'}={\bf b'}$$?

It turns out that (quite surprisingly to me) the answer is no: the degree $${\bf 0'}$$ is PA but not PA over anything not low.

Say that a degree $${\bf d}$$ is relatively efficiently PA if there is some $${\bf b}\le_T{\bf d}$$ such that $${\bf d}$$ is PA over $${\bf b}$$ and $${\bf d}$$ is low over $${\bf b}$$ (that is, $${\bf b'}={\bf d'}$$). By the observation cited above, $${\bf 0'}$$ is not relatively efficiently PA. However, in a precise sense "most" Turing degrees are relatively efficiently PA: the relativized low basis theorem tells us that the set of relatively efficiently PA degrees is unbounded, and so Martin's cone theorem (+ the fact that "relatively efficiently PA" is a sufficiently simple property) tells us that the set of relatively efficiently PA degrees contains a cone.

My question is roughly when this happens - that is, what might be a reasonable base for such a cone. Specifically:

Is every degree $$\ge_T{\bf 0''}$$ relatively efficiently PA?