DC and low signal Analysis of Common Source Amplifier: wrong drain current and amplification?
If I use your equations 

$$
I_d=(\frac{\sqrt{1+4*K*V_1*R_{SS}}-1}{2*\sqrt{K}*R_{SS}})^2 \\
V_1= (V_G -V_{SS})-V_{TO}
$$

And if I substitute the resistors in kiloohms and all the voltage in volts and all current in milliamps I get this: 

$$
I_d=( \frac{\sqrt{1+4*1.1*((-2.3077) - (-15) - 2)*5}-1}{2*\sqrt{1.1}*5} )^2 = 1.87719  
$$

http://www.wolframalpha.com/input/?i=((Sqrt%5B1+%2B+4*1.1*((-2.3077)+-+(-15)+-+2)*5%5D+-+1)%2F(2+Sqrt%5B1.1%5D*5))%5E2

And this result is also in milliamps.

\$I_D = 1.87719\textrm{mA}\$

The drain voltage is \$V_D = -3.77V \$

and \$g_m =2*\sqrt{K*I_D} = 2.87396\:\textrm{mS}\$

\$r_s = \frac{1}{g_m} = 0.347952 \textrm{k}\Omega = 347.952 \Omega \$

And the voltage gain is 

$$
\frac {V_{out}}{V_{in}}=-\frac {R_D ||R_L}{r_s+R_S||R_3} = \frac{5\textrm{k}\Omega}{0.347952\textrm{k}\Omega + \frac{1}{\frac{1}{0.2\textrm{k}\Omega}+\frac{1}{5\textrm{k}\Omega}}}=-9.25481 V/V
$$

Do you have any additional questions? And next time you should track your units with more care.