Conjugation in Fourier Transform
$X^*(j\omega)$ is the complex conjugate of $X(j\omega)$. So if

$$X(j\omega)=X_R(\omega)+jX_I(\omega)$$

then

$$X^*(j\omega)=X_R(\omega)-jX_I(\omega)$$

and

$$X^*(-j\omega)=X_R(-\omega)-jX_I(-\omega)$$

To gain some more understanding, consider the definition of the Fourier transform:

$$X(j\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt$$

So we have

$$X(-j\omega)=\int_{-\infty}^{\infty}x(t)e^{j\omega t}dt$$

and

$$X^*(-j\omega)=\int_{-\infty}^{\infty}x^*(t)e^{-j\omega t}dt=\mathcal{F}\{x^*(t)\}$$

as claimed in the quote of your question.