[d2cody]

Can someone answer the following question and briefly explain it? Thanks

[SVXX]

The more stable the conjugate base, the stronger the original acid.
I'd rank them as B > A > C. Here's my reasoning -:



I is the conjugate of A; II is the conjugate of B.

Let's start with A. Remove a proton from the alcohol group. The minus charge so formed can delocalize via resonance. The double bond shifts to form a minus charge on the other carbon of the double bond. This carbon is beta in proximity to the carbonyl group, and the minus charge is somewhat stabilized by its inductive effect.
Now for B. Following a similar process, the negative charge formed is on the carbon alpha in proximity to the carbonyl. Hence it is stabilized stronger by its inductive effect, than the previous case.
So B > A.
As C's conjugate base is not stabilized, it will be last.
Hence the order. If it's wrong, someone please do correct me.

[d2cody]

Thanks for the detailed answer   

[Agathiyar]

The order is B>A>C, in a alpha, beta unsaturated carbonyl compound beta carbon  is more electron deficient than alpha carbon, hence the hydroxyl at the beta carbon is more acidic than the hydroxyl at the alpha carbon. (another way of telling SVXX reply).