Hilbert transform linearity
Transform the signal into frequency domain

$$F(f) = \frac{1}{\left|a\right|} \left[~\frac{\Pi(f-\nu_{c})}{2} + \frac{\Pi(f+\nu_{c})|}{2} ~\right]$$

As you know the fourier domain representation of the hilbert operator is

$$H(f) = -j\text{sgn}(f)$$

Applying this transform means you just multiply $H(f)$ to $F(f)$. That gets you

$$\hat{F}(f) = -j\text{sgn}(f) \cdot \frac{1}{\left|a\right|} \left[~\frac{\Pi(f-\nu_{c}}{2} + \frac{\Pi(f+\nu_{c})|}{2} ~\right] $$

$$ \hat{F}(f) = \frac{1}{\left|a\right|} \left[~\frac{\Pi(f-\nu_{c})}{2j} + \frac{-\Pi(f+\nu_{c})|}{2j} ~\right] ~~ \text{Remember} -j = \frac{1}{j}$$

which is same as 

$$ \hat{F}(f) = \frac{\Pi(f)}{\left|a\right|} \ast  \left[~\frac{1}{2j} \delta(f-\nu) + \frac{-1}{2j} \delta(f+\nu) ~\right] $$

Inverse Fourier Transformation of this gives you

$$\hat{f}(t) = \text{sinc}(at)\sin{\left(2\pi\nu_{c}t\right)}$$

Hope this helps!