The MSSM has 2 complex doublets i.e. 8 real components in the Higgs fields. Four of them are electrically neutral (real bosons, antiparticles to themselves), four of them (i.e. two particle-antiparticle pairs) are electrically charged.
One neutral real boson and two charged ones (one charged pair) get eaten by the gauge bosons because 3 generators are broken, going from $SU(2)\times U(1)_Y$ to $U(1)_{\rm em}$.
The electrically charged ones are CP-partners of their oppositely charged antiparticles, of course. We don't learn anything if we consider CP for charged particles. One combination of the positive and negative particle is always CP-even and the orthogonal combination is CP-odd.
The four neutral ones are evenly split to two CP-even and two CP-odd ones, too.
It's the CP-odd ones that are eaten by the gauge bosons.
So one is left with the CP-even charged ones, three CP-even neutral bosons, and one CP-odd neutral Higgs boson. Why are CP-odd bosons eaten? It's because the gauge bosons pick a minus sign under C, simply because they're linked to generators which are C-odd, too. (Antiparticles need to be transformed by the opposite gauge transformation phases.) So under P, the gauge bosons are ordinary vectors, not axial vectors, but the C flips their sign, which is why they're "axial vectors under CP".
The Goldstone bosons that are eaten are CP-odd. Now, are they C-even, P-odd, or vice versa?
This is really a meaningless question because the spectrum of the electroweak theory (the fermions) isn't P-symmetric. Not even the free Lagrangian (not even when mixing is neglected). It's because there are left-handed doublets and right-handed lepton singlets, etc. So it makes no sense to try to assign P parity to fields - because no choice will lead to a P-invariant theory.
If you remove fermions, you may say that the Goldstone bosons are P-even, C-odd. They have the "normal sign" under P because the gauge bosons do, but they have the opposite one under C.
