Working out what happens to the brightness of a bulb in a simple circuit
This lead me to believe that if the current effectively doubles ... 


It won't double. With a true (ideal) voltage source as shown in the schematic the current through each branch will be independent of the presence or absence of any other branch. (If the source were a current source then the result would be different.)


  ... and the voltage stays the same then P = IV will mean that the power dissapated in Bulb A will also increase. 


Since V is constant and the R of the 'A' branch is constant (ignoring cold filament resistance on power-up) then I and P will be constant too. By substituting \$ I = \frac {V}{R} \$ you can rewrite your power equation as \$ P = \frac {V^2}{R} \$ and now it is clear that the power in each branch depends only on the supply voltage (a constant) and that branch's resistance.


  This means that Bulb A is now brighter? and the answer would therefore be C.


Oops!

If the question shows up this year go for 'B'.