what would the drain voltage be if NMOS source voltage is not ground?
The voltage at node Y is not what you say it is:

$$V_Y\ne V_{DD}-V_{THM1}-V_{THM2}$$

Rather, 

$$V_Y=V_{DD}-max\left(V_{THM1},V_{THM2}\right)$$

So, for \$V_{DD}=5\rm{V}\$, \$V_{THM1}=0.6\rm{V}\$, and \$V_{THM2}=0.5\rm{V}\$, then \$V_Y=4.4\rm{V}\$. M1 is in saturation with \$V_{DSM1}=0.6\rm{V}\$, while M2 is in triode with \$V_{DSM2}=0\rm{V}\$. 

If you flipped them around, so that \$V_{THM1}=0.5\rm{V}\$ and \$V_{THM2}=0.6\rm{V}\$, then they would both be in saturation. M1 would have \$V_{DSM1}=0.5\rm{V}\$, and M2 would have \$V_{DSM2}=0.1\rm{V}\$. \$V_Y\$ would still be \$4.4\rm{V}\$.

I should note that, since the body is disconnected from the S/D terminals, these MOSFETs will experience body effect. Therefore their threshold voltages will be substantially higher than we have considered in this analysis.