- Input Voltage = 16V
- Output Voltage = 5V
- Load Current = 0.1A
That means you want to convert (16-5) V · 0.1 A = 1.1 W to heat inside the IC.
The thermal data says that on a two layer board, you'll be heating up your IC by 1.1 W · 78 K/W = 85 K. I don't think you want that.
So, reduce the supply voltage with an external voltage regulator. To avoid producing the same heat elsewhere (and needing heatsinks and other expensive design constraints), use a simple buck converter.
Alternatively, use the external PNP transistor to get the heat production out of the IC, but honestly, that's more useful for having higher currents at a not-so-high input voltage than for absurdly high 16 V input voltage if you're still drawing 0.1 A. I mean, that's 1.6 W of power from the supply, just so you have 0.5 W at 5 V! What a wasteful design.
What are the ways to bring the output voltage of the under the undervoltage limit of the regulator?
Not quite sure what you mean, but undervoltage protection refers to input. So, you need an adjustable voltage supply.
We provide a step load at the output to see the voltage dip. But how much should the load current be and for how long?
You can't use the internal regulator alone because of the thermal considerations above, so you need to use an external PNP transistor as in Figure 6 of the datasheet.
Now, the U/V curve of the output will mostly depend on that, and your question can only be answered after you've chosen a PNP transistor, and knowledge on the control loop in the IC. I don't see any information on the control loop, only vague terms like "fast load ramping". You'll find a bit of information on how quick an undervoltage situation is detected in the "dynamic properties" table, but again, that's an input thing, not an output thing.
