[Dolphinsiu]

The following is Fischer projection     

     Ph                   Ph
      l                      l
Br ----  H           Br -----  H            OH^-
      l           ,          l                  ---->
H  ----  Ph         H  ----- Me
      l                      l
     Ph                   Ph

    (A)                  (B)

Which one is more reactive in elimination?

My guess is that B is more reactive as methyl group is electron-donating group which can stabilize the carbocation. Am I correct?

But someone say A is more reactive, as conjugated double bond is formed when Ph is attached, energy state become lower.

Which one is correct?

[Yggdrasil]

The second explanation clearly wrong because there is an sp³ carbon between the site of the carbocation and the Ph/Me.  Your first guess does make some sense.  However, what about E2 elimination?

[Dolphinsiu]

It is secondary halide! As strong base is used, E2 is favored.

But I still don't know when syn elimination is more favorable than anti-elimination, such as pyrolysis (syn elimination).

Do you know?