The potential difference across the bulb will be given by V=IR for the bulb.
As R is constant, the pd will depend only on the current flowing through it.
When you first connect the cell to the circuit there is no charge (and therefore no pd) on the capacitor.
As a result a current flows from the cell to the capacitor. The capacitor charges up and the pd across it increases. (V=Q/C)
When the capacitor is fully charged, the pd on it is equal to the emf of the cell and no more charge can flow. The current, in other words, falls off (exponentially) to zero.
As for the bulb. The current flowing through it, the charging current, is large to start with and the pd across it is equal to the emf of the cell. As the charging current falls to zero, the pd across it also falls to zero.
At the end there is no current flowing in the circuit so no pd across the bulb. All the pd is across the capacitor.