The need to block the flow of DC from the source to the load dictates the use of the matching network of the provided circuit image. But, first, let's get rid of the stray 40-pF capacitor by resonating it with a shunt inductor at 75 MHz .

$$
L=\frac{1}{\omega^2 C_{\text {stray }}}=\frac{1}{\left[2 \pi\left(75 \times 10^6\right)^2\left(40 \times 10^{-12}\right)\right]}=112.6 \mathrm{nH}
$$

This leaves us with the circuit shown in the provided circuit image. Now that we have eliminated the stray capacitance, we can proceed with matching the network between the 50-ohm load and the apparent 600 -ohm load. Thus,

$$
Q_s=Q_p=\sqrt{\frac{R_p}{R_s}-1}=\sqrt{\frac{600}{50}-1}=3.32
$$

$$
\begin{gathered}
X_s=Q_s R_s=(3.32)(50)=166 \mathrm{ohms} \\
X_p=\frac{R_p}{Q_p}=\frac{600}{3.32}=181 \mathrm{ohms}
\end{gathered}
$$


Therefore, the element values are:

$$
\begin{gathered}
C=\frac{1}{\omega X_s}=\frac{1}{2 \pi\left(75 \times 10^6\right)(166)}=12.78 \mathrm{pF} \\
L=\frac{X_p}{\omega}=\frac{181}{2 \pi\left(75 \times 10^6\right)}=384 \mathrm{nH}
\end{gathered}
$$


These values, then, yield the circuit. But notice that this circuit can be further simplified by simply replacing the two shunt inductors with a single inductor. Therefore,

$$
L_{\text {new }}=\frac{L_1 L_2}{L_1+L_2}=\frac{(384)(112.6)}{384+112.6}=87 \mathrm{nH}
$$