The saturated region is more complicated than shown in that diagram. CL's comment says that it 'actually goes down by a little bit', which is true for low base voltages, but then at higher base voltage it does go up again.
The initial drop of 'saturation' voltage is where it goes from 'soft' saturation to 'hard' saturation. There is no well-defined distinction between these two, it's simply the ratio of collector to base current. As the ratio falls below hFE, the saturation becomes 'harder', and the saturation voltage falls a little more.
As the diagram has been drawn with straight lines, we can assume the exercise is asking you to ignore this feature of the transistor, and treat saturation as if it happens at a unique base voltage, rather than be smeared out a little as discussed above.
As the base voltage increases further, the base current increases further. The base-emitter diode is now turned on fully, and the residual resistance of the emitter is limiting the base-emitter current. Current through the emitter resistance lifts the emitter voltage, and with it the collector voltage, as shown approximately in the graph.
I must say I find it a bit odd to show a rising straight line. The simplest 'ideal' model is a flat straight line. If we are going to include further accuracy, then I think it would be appropriate to include both the dip and the rise. However, the graph in the exercise has shown only the rise, and that's the question you're being asked to answer. I suppose the model can be extended by the addition of a simple linear emitter resistor to explain this effect. However, the fall before it has more complicated physics. I am assuming this question was asked following a presentation on transistor models.