Edit: This answer is based on a typo in the original question referring to a different starting compound than the OP intended.
A quick analysis of this reaction gives me the following:
3-ethyl-2-methylpentan-1,2-diol can be protonated by acid to produce cation A or B. Either of these cations can lose water to form a carbocation
- The carbocation formed from A is a tertiary carbocation C that might not be likely to rearrange, but we'll consider the possibility in a bit.
- The carbocation formed from B is a primary(!) carbocation D that seems unlikely to form.
Pinacol rearrangement from cation A
A pinacol rearrangement from cation A through carbocation C forms an aldehyde: 3-ethyl-2-methylpentanal.
Pinacol rearrangement from cation B
Having established that carbocation D is unlikely to form, we might suspect no rearrangement, but the rearrangement can occur in a concerted fashion with the loss of water. The end result is a ketone: 4-ethyl-3-hexanone. These kinds of concerted rearrangements to avoid primary cations are also known to occur in Friedel-Crafts reactions.
Neither product seems to match what you provided as the book's answer. Is it possible you miss-named the product?
Which product is more likely the major product?
Rather than tell you, I will give you a hint:
One rearrangement produces a thermodynamically more stable cation than the other.