Answer
Bulb A gets brighter, and bulb B goes out.
Work Step by Step
Let $R_A$ be the resistance of bulb A.
Let $R_B$ be the resistance of bulb B.
When the switch is open, the current is $I = \frac{V}{R_A+R_B}$
The power dissipated by bulb A is $I^2~R_A = (\frac{V}{R_A+R_B})^2~R_A$
The power dissipated by bulb B is $I^2~R_B = (\frac{V}{R_A+R_B})^2~R_B$
When the switch is closed, all the current flows through the section of the circuit where the switch has just been closed since there is essentially no resistance there. The current is $I = \frac{V}{R_A}$
The power dissipated by bulb A is $I^2~R_A = (\frac{V}{R_A})^2~R_A$
The power dissipated by bulb B is $I^2~R_B = (0)^2~R_B = 0$
The brightness of the bulb depends on the power that is dissipated. Therefore, when the switch is closed, bulb A gets brighter, and bulb B goes out.
