How to determine the stereochemistry after elimination of anti-1-bromo-2-methylcyclohexane?
You diaxial intermediate is good and you can clearly see that there is no way to achieve a Zaytsev product in an $\mathrm{E2}$ elimination since that would require eliminating the methyl group. Carry on with the elimination and then you have a double bond in the top half of the molecule and the methyl group pointing downards — molecule E if inverted by a vertical plane of symmetry traversing the $\ce{C-Me}$ bond.

But the product molecules are drawn with the double bond on the bottom. Thus, you need to rotate the molecule by $180^\circ$. If you do this with a rotation axis in the paper plane (again one parallel to the $\ce{C-Me}$ bond), the double bond is in the correct position but the methyl group is pointing upwards.

You can verify this by determining the absolute configuration of the ipso-methyl carbon: it should be (R) according to your chair drawing and it is (R) in answer C.