It's really a shame you couldn't have provided some of that information earlier. I'm not sure if it affects any answers I've given, but it certainly would have helped provide more certain answers.
I'm no sure you can say R200 and TM+ act as a voltage divider. It all depends on what's connected to TM+ -- is it any more complex than a thermistor to ground?
It begs the question of whether the voltage at TM+ can ever exceed Vdd5, but I'll let you sort that out.
The first sentence of the third paragraph doesn't really make sense. Perhaps you mean the input to the microcontroller? If this was the sole reason, there are far easier ways of protecting the input (the dual diode and a resistor would be sufficient)
You continue using T201 for two devices. You should use T201a and T201b.
The term "dummy load" in paragraph 3 is incorrect.
The 4th paragraph is written in a hard-to-understand manner. What happens when the voltage is in a normal range? 12V is not the magic voltage. It is Vdd + the turn on voltage of T201a.
The last sentence is correct.
12 V is just an example that I have taken. It could be anything ranging from 6 to 13.5 V.
What should I use instead of the word "dummy load" ?
In the normal condition, none of the transistors turn on. The voltage at point K1A2 will always be less than 5V and this voltage will be available at output 11G and the microcontroller pin will be safe.
Let me take TM+ as 8 V, so the voltage at point K1A2 will be 8V too, but at point P204 , it will be a little less say 7.3. , since the Veb= 7.3-5 = 2.3 >0.7 , the transistor saturates , is this analysis correct ?
suppose the transistor is saturated, it behaves like a closed switch but still there will be some voltage drop across it , say 0.2V (in general). so now what is the voltage that is available at point P203 ?