Why is the initial carbocation not formed adjacent to the phenyl moiety?
I agree with your sentiment, so I run a very quick calculation with a semi-empirical method (GFN-xTB) to find the protomers. This has been done using water as solvent. Note that I did not choose to include any deuterium.

First I calculated the conformational space of the plain molecule (NIST: 1,1'-(3-methyl-1-propene-1,3-diyl)bis-benzene, InChI=1S/C16H16/c1-14(16-10-6-3-7-11-16)12-13-15-8-4-2-5-9-15/h2-14H,1H3/b13-12+, Google search InChIKey: GNQWHYWLSGTMSL-OUKQBFOZSA-N) to obtain the lowest lying conformation and show you the HOMO:



As you can see the double bond is a good protonation cite, with both carbon atoms having about equal contribution; one would therefore expect that the most stable intermediate will form. Undoubtedly this should be the one where the charge can be delocalised into the aromatic &pi;-system.

The automatic generation of the protomers agrees here, and the lowest lying conformation (with partial &pi;-stacking) is shown below, the protonated carbon indicated in bright white:



There is a tentative pathway via a four-membered (green) transition state to rearrange to the product shown in the exam question, but I do not have the time to investigate. Therefore here is my proposed mechanistic scheme: