Which cannot happen, because the supply is the highest voltage you have
So, then you need to connect the source to the load
But, say your FET needs 3V of gate voltage to turn on
And your gate can only go to the supply, which is 5V
Then the source can only go to 2V, because if it goes higher the FET will turn off
So the load will only get 2V
Say the load is another FET structure of the same type
It can now no longer turn on, because the gate never goes above 2V and it only turns on at 3V
Adding in the bode diode adds another problem with the first scenario, where the source connects to the supply
Because the bode diode in a FET, which is a part of how it works, not an option, conducts positive current from the source to the drain
So connecting the source to the supply will make it always be on, because the body diode will conduct
The same reasoning works for P-FETs on the Ground/Negative-supply
You could say "We'll make chips that work on 5V, but also get a 12V to turn on all the FETs on the positive rail" and such chips do exist. In fact, many earlier types of electrically erasable chips needed 12V to program or erase exactly for that reason.
But, since we have P-FETs and N-FETs it's infinitely cheaper to use complementary devices and supply everything from one supply voltage, both inside the chip and outside it, than having to plan extra positive or negative supply lanes
And now that our technology has become better, they also make the same kinds of memory as used to need a high voltage for erase that can work with the normal single supply
Because even if it needs a bit more detailed work in the chip, the cost of having various supplies in systems and the chip itself is bigger
