What would be the product (B) in the following sequence of reaction? Here is how I approached this:
As we know, alc. KOH results in dehydrohalogenation of the compound which takes place as an anti elimination. And then, we're simply conducting halogenation of the compound which would result in a dibromo substituded product.

Where I'm facing problems is deciding the final orientation of the product. How will we represent the product in the fischer representation as the initial reactant is given in its fischer representation as well.

For reference, here's the product (B) given in my book:
This is a fun one, and a common question in this course.  First, as you say, H and Br are lost by anti elimination, since they helpfully tell you base and E2 are involved.  That means the H and Br are lost from opposite faces of the future double bond, leaving us with a trans intermediate that has a mirror plane (the plane of the molecule)

Now we need to add Br+ to make a bromonium intermediate.  Notice that both carbons end up as (R) or else both end up as (S), so two enantiomers are possible here.

But, whichever carbon is attacked by Br- ends up with an inversion of symmetry, so you now have two carbons of opposite chirality.  The result, as shown in your answer, is a molecule with a mirror plane.  In other words, adding bromine to opposite sides of a trans double bond makes both sides come out as mirror images, once you turn the bromines around to line up.  This compound is a meso compound.  But ... what about the other isomer if the Br+ originally goes to the other side of the double bond?  Well, just turn your Fisher structure upside down. :)