What happens to the brightness of bulb a when the switch is closed and bulb b lights up?

The problem describes the relationship of "bulb a" and "bulb b" to be in connected in series. When the switch is open then no current can flow, on the other hand, when it is closed, current will pass through.

When only "bulb a" is connected to the battery then more current is flowing to "bulb a" causing it to be bright.

Closing the switch would mean that "bulb b" is already included in the circuit and the battery will push small current to flow around the whole circuit. The more bulbs are connected, the harder for the current to flow because the resistance will be very high.

So the light of "bulb a" will be dimmer.