While the scheme posted by discodermolide could be the mechanism, I think it is very unlikely. If it were true, then a) should be the correct answer. I see no reason for a hydride shift to form the carbocation intermediate. If it were more stable than the initial carbocation, then the option of forming it directly should be a faster reaction. For example, dehydration of 3-methyl-2-butanol can give a tertiary carbocation intermediate as it is more stable than the secondary carbocation. If you compared rates, I'm sure the tertiary alcohol could be dehydrated faster than the secondary.
I think you should approach this problem differently. If you know that c) is the correct answer, then what is the rate determining step? For most alcohols, the rate determine step is the formation of the carbocation intermediate. I don't think that is the case here or the 4-OH should be the fastest. The rate determine step must be the loss of the proton. Whether this occurs from the enol of the ketone or directly from the ketone, I don't know.