Answer
See the detailed answer below.
Work Step by Step
We know that
$$\phi=\tan^{-1}\left[\dfrac{X_L-X_C}{R}\right]$$
Let's draw than in a right triangle, as shown below.
Now we can see that
$$\cos\phi=\dfrac{R}{H}\tag 1$$
We can see, from the figure below, that the hypotenuse of the triangle is
$$H=\sqrt{(X_L-X_C)^2+R^2}=Z$$
where this is the impedance $Z$, so $H=Z$.
Plug that into (1),
$$\boxed{\cos\phi=\dfrac{R}{Z} }$$
$$\color{blue}{\bf [a]}$$
We know that the current in an $ RLC$ circuit is given by
$$I=\dfrac{\varepsilon_0}{Z}$$
where $\varepsilon_0=I_{\rm max} R$, so
$$I=\dfrac{I_{\rm max} R}{Z}$$
where, from the boxed formula above, $R/Z=\cos\phi$
$$\boxed{I= I_{\rm max}\cos\phi}$$
$$\color{blue}{\bf [b]}$$
We know that the source power is given by
$$P_{\rm source}=I_{\rm rms}\varepsilon_{\rm rms}\cos\phi$$
where $I_{\rm rms}=I /\sqrt{2}$, and $\varepsilon_{\rm rms}=\varepsilon_{ 0}/\sqrt{2}$, so
$$P_{\rm source}=\dfrac{ I}{\sqrt{2}}\dfrac{\varepsilon_{ 0}}{\sqrt{2}}\cos\phi$$
Plug $I$ from the boxed formula above,
$$P_{\rm source}=\dfrac{ I_{\rm max}\cos\phi}{\sqrt{2}}\dfrac{\varepsilon_{ 0}}{\sqrt{2}}\cos\phi$$
$$P_{\rm source}=\overbrace{ \frac{1}{2}I_{\rm max} \varepsilon_{ 0}}^{P_{\rm max}}\;\cos^2\phi$$
$$\boxed{P_{\rm source}= {P_{\rm max}} \cos^2\phi}$$
