Li-po battery protection
I'm sorry, but your schematic will not work this way:


The AS393P-E1 is a comparator. That means it has an open collector/drain output. You need to have a pull-up resistor at the output, otherwise the gate of your transistor will be floating when the output of the comparator is high. Not good.
The Gate-Source Threshold Voltage of the BS250P-DIC is max. -3.5V@Id=-1mA. So even applying the full battery voltage of 3.6V to the gate isn't going to be enough to power your load properly. 
The voltage divider is slowly draining your battery (12uA).


What do you mean with 'I want to have reserve'? Is this an additional circuit, after the circuit that's built into the battery? If not, and this is the only protection circuit, go with the suggestion of @Marcus. Then please be careful: Li-Po batteries can really explode or burn when not handled properly!



Edit:

Ok, I promised you a 'keep it simple' answer. Here's a schematic, based upon the schematic you posted.

I used the MCP1501T-10E/CHY you suggested. It outputs 1.024V. The 1.69M/1M voltage divider outputs 1.024V when the battery is at 2.755V. I also added pullup resistor R4, to force Q1 OFF when the battery voltage is too low. I kept the comparator you suggested. Please note that the Source and Drain of the transistor in your schematic were wrong as well. I only saw that just now. So make sure you wire the transistor the correct way, as in my schematic.

Now, there are a few problems:


I couldn't find a P-channel enhancement MOSFET (Transistor Q1), that has a low enough Gate Threshold Voltage at the required 75mA. Maybe someone else on this website knows one? Problem is also that I don't know where you're 'shopping' for one. But I don't want to invest any more time in searching. I'm sorry.
This circuit consumes power, as I already mentioned. Problem is that it continues to consume power when the battery is near empty. So that's not a very good protection I guess. It consumes about 1uA for the voltage divider + 140uA for the voltage reference + 600uA for the comparator = 741uA. It's not super much, but enough to completely drain you battery in time. So be careful there.


Also take a look here. It has some really good solutions, that might be much better then the schematic I've posted. Especially the 'yellow' schematic seems like something you might fancy.

Looking at your comments, I got the impression that you wanted a schematic similar to what you posted, so I hope this helps.



Update:

This is the new schematic.


As you can see, I did NOT use the TL431, as it has a minimum output current of 1mA, which is way too much for your purpose. So that will not work. I used a LM385Z-1.2G instead, which has a minimum output current of 10-20uA - way better. It outputs a voltage of 1.235V

So how to calculate the resistors?


R9 needs to make sure that the minimum current is 20uA. Let's say we want the regulator to work upto 2.235V. That's 1V dropout (2.235-1.235=1V). R=U/I, so R=1V/20uA = 50k. 50k is a hard-to-find value, as resistors go in E-ranges. 47k is in the E6 range, so very easy to find. (The higher the E number, the higher the precision (lower tolerance), and the harder it gets to find + more expensive).
E6 means there are 6 values per decade (6 values for 0-9, 6 for 10-99, 6 for 100-999, ...)
E12 means there are 12 values per decade
You have E6, E12, E24, E48, E96, E192


R5 / R6:

This is a voltage divider. The input current of the opamp is very low, so we can leave that out for simplicity.


The voltage Vout = Vin * R6/(R5+R6). You can even use an online calculator.
Vin is the Voltage at which you want to switch off your battery: 3V
Vout is the voltage of your voltage reference: 1.235V
You assume one resistor, and calculate the other. The current that will flow is I=U/R =Vbat/(R5+R6), so you don't want the resistors too small, as the power consumption will be too high. You also don't want them too large, as your schematic will be more subject to interference. Probably R5+R6 somewhere between 500K and 2M will be fine.
Let's assume R6 (use a value from E12 or E24). That gives you a value for R5. Find the value closest in E12 or E24, and calculate the dropout Voltate Vin again with the right resistor values. Go to step 4, and repeat, until you find a good combination that comes close enough to what you want. Use values from E12 or E24, as they are usually easy to find.


I found a good combination for R5=1M, and R6=680k. That gives you a cutoff Voltage of 3.051V - close enough. Another option would be R5=1.5M and R6=1M. That gives a cutoff of 3.088V, also close enough I think. With 2.5M total resistance, your power consumption would be less, but both are negligible compared to the power usage of your comparator anyway. The 2.5M option would give a bit more chance of interference, so I picked the lower option.

The process of 'assuming' one resistor to calculate the other is a bit hard in the beginning, but as you use components more, and learn the E values by heart, it gets easier. Capacitors and the like also use E ranges, so you'll use them a lot!

Good luck!