Solving Currents in BiCMOS Darlington Pair
Well for this circuit we have 
\$I_D + I_C = \frac{V_{CC}-V_{Out}}{R_L} = 1mA\$ 
Additional we knows that \$I_S=I_D=\frac{V_{BE}}{R_B}\$ 
So we we assume \$V_{BE} = 0.6V\$ we have \$I_D= 0.6mA\$ and \$I_C=0.3mA\$ therefore \$V_{BE} = V_T*ln\left(\frac{Ic}{Is}\right)= 0.7469V \$ (I assume Vt = 26mV). 

So we have a new value for Vbe, so, the new value for Id and Ic is: 

\$I_D = \frac{0.7469V}{1k} = 0.7469mA\$ 

\$I_C = 0.253mA \$ 

so again we can find new value for \$Vbe\$ 

\$Vbe =V_T*ln\left(\frac{Ic}{Is}\right)= 26mV *ln(\frac{0.253mA}{0.1fA}) = 0.74254V \$  and the new \$I_D = 0.74254mA ;I_C = 0.25746mA\$ values. 

And once more I repeat the iteration 
\$Vbe = 26mV *ln(\frac{0.25746mA}{0.1fA})=0.742995V \$ 

\$I_D = 0.742995mA ;I_C = 0.257005mA\$

The new \$Vbe\$ value is \$Vbe = 0.742949V\$

At this step, I end the iteration process and conclude that \$Vbe = 0.7429V\$.

And \$I_D = 0.7429mA\$ and \$I_C=0.2571mA\$ 

Since we know the MOS drain current \$I_D\$ we can find \$Vgs\$

\$V_{gs} = V_{th}+\sqrt{\frac{I_D}{0.5k}} = 0.6V+\sqrt{\frac{0.7429mA}{0.5m}} =1.81893V\$  

And finally \$V_{BIAS} = V_{BE}+V_{gs} =0.7429V+1.81893V = 2.56183V\$

In all this, calculations I ignore the BJT base current. 

EDIT

To get exact solution you need to solve this:

$$I_C = 1mA - \left(\frac{I_C}{\beta}+\frac{Vbe}{1k}\right);I_C = 1*10^{-16}*e^{\frac{Vbe}{V_T}}$$

And if I plug this into computer I get \$ V_{BE} =0.742718V; I_C=0.254735mA \$