PMOS Circuit Problem
From \$V_{SD} = 7.5V\$ we can find \$R_S = \frac{15V - 7.5V}{3mA} = 2.5\:\textrm{k} \Omega\$ 

The PMOS gate-source voltage \$V_{GS}\$ at \$I_D = 3\:\textrm{mA}\$ 

Can be found from this equation: 

$$I_D = \textrm{K}(V_{GS} - V_{TH})^2$$ 

solving this for \$V_{GS}\$ we get this:  

$$V_{GS} = \sqrt{\frac{I_D}{K}} + V_{TH} = \sqrt{\frac{3mA}{0.2mA/V^2}} + 2V = 5.873V $$

So, the gate voltage needs to be \$5.873\$ volts lower than the source voltage. 

$$V_G = 7.5V - 5.873V = 1.627V $$  

Therefore you need a voltage divider that will provide this voltage to the PMOS gate.  

So we have: 

$$V_G= V_{DD} \cdot \frac{R_{G2}}{R_{G1}+R_{G2}} = 1.627V$$

$$ R_T = \frac{R_{G1} \cdot R_{G2}}{R_{G1}+R_{G2}} = 3\textrm{M}\Omega $$

Solving this we have 

$$R_{G2} = R_T \cdot \frac{\frac{V_{DD}}{V_G}}{\left(\frac{V_{DD}}{V_G} - 1\right)} \approx 3.3\textrm{M} \Omega$$

$$R_{G1} =R_{G2}\cdot \left(\frac{V_{DD}}{V_G} - 1\right)\approx 27\textrm{M}\Omega $$

As for the part \$C\$

$$C1 &gt; \frac{1}{2 \pi F\cdot R} = \frac{1}{2 \pi 20\textrm{Hz}\cdot (10\textrm{k}\Omega +3\textrm{M}\Omega)} \approx \frac{0.16}{20\textrm{Hz}\cdot 3\textrm{M}\Omega} \approx 2.7\textrm{nF}$$

$$C_2 &gt; \frac{0.16}{20\textrm{Hz} \cdot (2.5\textrm{k}\Omega+1\textrm{k}\Omega)} \approx 2.2\mu\textrm{F}$$