The current flowing through RD is

$I_D=\dfrac{V_{DD}-V_D}{R_D}$

$I_D=\dfrac{14-8}{1.6\times10^3}$

$\therefore \underline{\underline{I_D=3.75\ mA}}$

Applying KVL to the Drain - Source Loop

$V_{DS}=V_{DD}-V_D$

$V_{DS}=14-8$

$\therefore \underline{\underline{V_{DS}=6\ V}}$

Using the Shockley Equation to find VGS

$I_D=I_{DSS}\bigg[1-\dfrac{V_{GS}}{V_P}\bigg]^2$

$3.75\times10^{-3}=8\times10^{-3}\bigg[1-\dfrac{V_{GS}}{-4}\bigg]^2$

$3.75\times10^{-3}=8\times10^{-3}\bigg[1+\dfrac{V_{GS}}4\bigg]^2$

$3.75\times10^{-3}=8\times10^{-3}\big[1+0.25\ V_{GS}\big]^2$

$3.75\times10^{-3}=8\times10^{-3}\big[1+0.5\ V_{GS}+0.06\ {V_{GS}}^2\big]$

$3.75\times10^{-3}=8\times10^{-3}+4\times10^{-3} V_{GS}+0.5\times10^{-3}{V_{GS}}^2$

$0.5\times10^{-3} \ {V_{GS}}^2+4\times10^{-3}\ V_{GS}+4.25\times10^{-3}=0$

$\therefore V_{GS}=-1.26\ V\ \&\ -6.73\ V$

But $V_{GS}\not= -V_P$

$\therefore V_{GS}=-1.26\ V$

Applying KVL to the Gate - Source Loop

$V_{GS}=-V_{GG}$

$\therefore \underline{\underline{V_{GG}=1.26\ V}}$