good morning again

I have some doubts about what was happening in this example case.
I have 1-phenyl-2-iodocyclohexane in with potassium tert butoxide. It is clear that the idea converges by an E2 mechanism. However, I have several doubts.
I have drawn conformations A and B, the first thing is that it is displaced towards conformation B since there is less 1,3 diaxial interaction since phenyl is much more voluminous than iodine and fits better in the equatorial zone.
At this point, in conformation B, there are 2 plausible elimination hydrogens, the germinal one with phenyl and the neighboring one with iodine, I hope I make myself clear, I have circled them in green.
Here my question arises, which is easier to withdraw? normally the more substituted alkene is formed, however I wonder if that phenyl group won't generate some steric hindrance and eventually the other one will be removed? to give the Hoffman product. This would be my main question
Then I have a secondary doubt, which had arisen at the beginning, but I have already realized by observing the conformations better. It is about the possibility of leaving the phenyl group instead of the iodine, since the phenyl in the A conformation also has an antiperiplanar hydrogen. But hey, I discard this point because the conformation is not favored.
However, the question would be if, in the right circumstances (not in this case) could phenyl be a leaving group? I have not yet had to study aromatic compounds or the stability of phenyl as a cation.
A greeting and a thousand thanks in advance