I made this bromoacid (alpha-bromo-4-phenylbutyric acid) using Br2, 4-Phenylbutyric acid & PBr3 (cat.) using this procedure : http://www.orgsyn.org/orgsyn/prep.asp?prep=cv1p0115
Here I think, the alpha-bromination goes easily only to COOH according to this route (Not sure).
First, I guess you mean 'alpha to COOH' and not 'to COOH'. And why aren't you sure?
Concerning your question, very few organic reactions are strictly
selective, yielding just one product with absolutely no interference. We usually rely on carefully chosen conditions to prevent side reactions from occurring.
In your case, having PhCH2CH2CH2COOH and Br2 around could make you think of at least two competing reactions:
1 - aromatic electrophylic halogenation (on Ph)
2 - free-radical benzylic halogenation (on the CH2 attached to Ph)
However, for side reaction 1 you would need some Lewis acid to activate the halogen. I think PCl3 or PBr3 are not going to do that, at least not as efficiently as AlBr3, AlCl3...
For side reaction 2 you need atomic bromine, which is generated by the action of UV light on Br2. Maybe you want to avoid exposing your reaction flask to sunlight or to strong artificial light (a good old aluminum foil around it will do the job).
Anyway, I think orgsyn describes some procedures of HVZ bromination on n-phenyl-carboxylic acids. If competing processes mean you get a lower yield, well, either accept it or try a different route...