Weakening of the perfect set property

The perfect set property says that every uncountable set of reals contains a perfect subset.

Now consider the following statement:

P: For every $$X\subset\mathbb{R}$$, either $$X$$ or $$\mathbb{R}\setminus X$$ contains a (non-empty) perfect set.

Clearly, the perfect set property implies the second one. I've heard that the converse is false, but I couldn't find a model of ZF+P (this would obviously entail a failure of choice). Moreover, is it possible to get a model where P holds without assuming the existence of an inaccessible?

Any references for this? I've looked at Jech, Rubin, this paper , and Googled a bunch, but found nothing.

Specker's theorem tells us that the perfect set property implies $$\omega_1$$ is a limit cardinal in $$L$$, but without countable choice, $$\omega_1$$ can be singular. And Truss shows that this is consistent with the perfect set property.
Moreover, he shows, in a fairly simple family of models every set of reals can either be well-ordered or it contains a perfect set (and the reals cannot be well-ordered), which is an intermediate statement between your $$\rm P$$ and the perfect set property. This is even consistent with Dependent Choice to some fixed level.