Finding Drain voltage of current mirror
This is a very easy task. You already should now that in the current mirror 

\$I_{D1} = I_{D2} = I_{D3} = I_{D4}\$

And this can only be true if all transistor have the same \$V_{GS}\$

Therefore the current through the \$R_1\$ resistor is 

\$I_{R1} = I_{D1} = \frac{20V - 2V_{GS}}{R_1}\$ 

Hence the equation needed to solve for \$V_{GS}\$ is:

$$\frac{20V - 2V_{GS}}{R_1} = \frac{K_p}{2}(V_{GS} - V_T)^2$$ 

And the solution is \$V_{GS} \approx 1.05\textrm{V}\$ and the current is \$I_{R1} \approx  124.3\mu \textrm{A}\$ 

This picture tries to explain why I use \$2V_{GS}\$



Also, notice that from KVL we can wite  

\$I = \frac{V_{DD} - V_B}{R}\$ 

And in your circuit \$ V_{DD} = 15V + 5V = 20V\$