Protecting ADC pin from the opamp output
Figure 1. The OP's circuit.

Here are a few comments:


This looks suspiciously like an LM317 regulator in current control mode. That means your output voltage could swing from 0 (in a short-circuit) to about 9 V. You are showing a variable resistor in there which suggests that you are thinking of making it adjustable. I suggest you calculate the current you are going to run through this and then start to look at the current rating of typical potentiometers. Remember that the power rating for a pot assumes dissipation over the full length of the resistance track. If you have it set to mid-position then the power rating is halved. (i.e., work out the maximum current the pot will handle.)
You have several parallel loads. Since these are driven by one constant-current source you won't be able to control the current in each branch. Your shunts are on the "high side" of the loads and so could see up to 9 V on them.
This is OK and will switch off all loads.
Your multiplexer will have to cope with the 9 V on the shunts. You have no supply voltage indicated for this device so this may be the first to blow.
Your op-amp has no gain control and no offset adjustment to remove the high-side voltage.


It would help if you edit your question to explain what exactly you are trying to make and why.

Low-side current monitoring simplifies everything as the multiplexer and op-amp can be powered from the 5 V microcontroller supply. 

What is the purpose of the MOSFET (3)?