Why is the initial carbocation not formed adjacent to the phenyl moiety? I know that after rearrangement the carbocation is more stable, but we should protonate the compound first where the carbocation it is more stable, i.e.  adjacent to the phenyl group. Why do they suggest a different pathway? Is the shown correct?
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: