Prove a property using shift theorem and duality
If $X(f)=\mathcal{F}\{x(t)\}$ you have, in analogy with Eq. $(7)$,
$$x(f)\Longleftrightarrow X(-t)\tag{1}$$
where we assume that functions with the independent variable $f$ are Fourier transforms of the corresponding functions with independent variable $t$.
Applying $(1)$ to
$$x(t-b)\Longleftrightarrow e^{-2\pi jbf} X(f)\tag{2}$$
gives
$$x(f-b)\Longleftrightarrow e^{2\pi jbt} X(-t)\tag{3}$$
According to $(1)$, $X(-t)$ is the inverse Fourier transform of $x(f)$, so with $g(t)=X(-t)$ and $G(f)=\mathcal{F}\{g(t)\}=x(f)$ you obtain the desired result:
$$G(f-b)\Longleftrightarrow e^{2\pi jbt} g(t)\tag{4}$$