What happens when prop-2-en-1-ol reacts with concentrated hydrogen bromide
Rather than trying to rationalise what the question wants to know, I ran a quick SciFinder search. One of the first results was the following reaction, reported for the perdeuterated allyl alcohol by Thiem, Mohn and Heesing in 1985:[1]


Scheme 1: Reaction of perdeuterated allyl alcohol with $\ce{HBr}$. All implicit hydrogens in this scheme are deuterium atoms.[1]

Thiem, Mohn and Heesing do not go into details regarding the mechanism, but as I commented it really doesn’t matter whether the reaction follows a $\mathrm{S_N1}$, $\mathrm{S_N2}$ or $\mathrm{S_N2'}$ mechanism, since the final product will always be the same. In any case, the first step will be protonation of the hydroxy group to turn it into a better leaving group.

Another one of the first results was the following, reported by Karki and Magolan in 2015:[2]

Scheme 2: Reaction of normal allylic alcohol with $48~\%\ \ce{HBr}$ in DMSO/chloroform.[2]

Their proposed mechanism includes the in situ creation of $\ce{Me2S^+-Br}$ which will then react with the double bond giving a bromonium ion; this will then be captured by a second bromide to give 2,3-dibromopropan-1-ol.

I was unable to find single-step reactions of allyl alcohol with hydrogen bromide leading to either 2-bromopropan-1-ol (which would imply a single Markovnikov addition of $\ce{HBr}$ onto the double bond), 1,2-dibromopropane (which would imply Markovnikov hydrobromination of the double bond of allyl bromide) or 1,3-dibromoprop-1-ene (where I have no clue how exactly that would be generated from allyl alcohol in the first place). Funnily enough, you state that this last product, C according to your scheme be the correct answer. I challenge the exam.



References:

[1]: J. Thiem, H. Mohn, A. Heesing, Synthesis 1985, 775. DOI: 10.1055/s-1985-31344.

[2]: M. Karki, J. Magolan, J. Org. Chem. 2015, 80, 3701. DOI: 10.1021/acs.joc.5b00211.