Answer: Step 1 ) Calculate $V_{GS}$

Since it is given that $V_P=-2V \space and \space I_{DSS}=4mA$

Now, $V_{G}=\dfrac{R_2}{R_1+R_2}V_{DD}$

$V_G=\dfrac{8.57}{12+8.57}\times 24$

$\therefore V_G=9.9V$

Now, $V_{GS}=V_G-V_S$

$V_{GS}=V_G-I_DR_S$

$\therefore V_{GS}=10-3I_D$

Step 2) Calculate $I_D$

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

$I_D=4[1-\dfrac{(10-3I_D)}{-2}]^2$

$4I_D=4[2+10-3I_D]^2$

$I_D=144-72I_D+9I_D^2$

$9I_D^2-72I_D+144=0$

$\therefore I_D=\dfrac{72}{18}=4mA$

Step 3) Calculate $V_{GS}$:

$\therefore V_{GS}=10-3(4)$

$\therefore V_{GS}=V_P=-2V$

$V_{DS}=V_{DD}-I_D(R_D+R_S)$

$\therefore V_{DS}=24-4 \times 10^{-3}[910+3 \times 10^3]$

$\therefore V_{DS}=8.36V$

With increase in $V_{DS}$ there will be no futher increase in drain currentbeacause it has already reached its maximum possible value i.e. $I_{DSS}$ . This shows that the device is in the pinch off region