Since the MOSFET is to be in saturation, we can write

$$
\begin{aligned}
I_D & =\frac{1}{2} k_P^{\prime} \frac{W}{L}\left(V_{G S}-V_t\right)^2 \\
& =\frac{1}{2} k_p^{\prime} \frac{W}{L} v_{O V}^2
\end{aligned}
$$


Substituting $I_D=0.5 \mathrm{~mA}$ and $k_p^{\prime} W / L=1 \mathrm{~mA} / \mathrm{V}^2$ and recalling that for a PMOS transistor $V_{O V}$ is negative, we obtain

$$
V_{O V}=-1 \mathrm{~V}
$$

and

$$
V_{G S}=V_t+V_{o V}=-1-1=-2 \mathrm{~V}
$$


Since the source is at +5 V , the gate voltage must be set to +3 V . This can be achieved by the appropriate selection of the values of $R_{G 1}$ and $R_{G 2}$. A possible selection is $R_{G 1}=2 \mathrm{M} \Omega$ and $R_{G 2}=3 \mathrm{M} \Omega$.

The value of $R_D$ can be found from

$$
R_D=\frac{V_D}{I_D}=\frac{3}{0.5}=6 \mathrm{k} \Omega
$$