It depends on which degrees of freedom are entangled, and how the absorption and emission process works exactly. It is certainly possible to design this in such a way that case a) happens -- A and the re-emitted B are still entangled, in exactly the same way as A and the original B photon were.
For instance, https://www.nature.com/articles/nature11023 reports experiments where the state of a photon is mapped onto an atom, and conversely, the state of the atom is mapped back onto an emitted photon -- such a setup would do precisely what I describe above: The state of the absorbed photon (which includes any kind of entanglement of B, due to the linearity of quantum theory) would be stored in the atom, and subsequently be mapped back onto the re-emitted photon.
On the other hand, c) can also happen -- in fact, it is always an accurate description, regardless of the process, if you only look at photon B. Then, however, the information carried by the entanglement should be gone somewhere, meaning that the photon is still entangled with something, just not the new photon A.
b), on the other hand, never happens. In fact, I don't know what a "superposition of all eigen states of linear polarisation" would precisely be.