Solution

The pair of linear equations
$$ 13x+ky-k=0$$
$$ 39x+6y-(k+4)=0$$

Here, $$a_{1}=13,b_{1}=k,c_{1}=-k$$
$$ a_{2}=39,b_{2}=6,c_{2}=-(k+4)$$

The system has infinitely many solutions if
$$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$$

$$\Rightarrow \displaystyle \frac{13}{39}=\frac{k}{6}=\frac{-k}{-(k+4)}$$

Taking,
$$ \displaystyle \frac{13}{39}=\frac{k}{6}$$

$$\Rightarrow \displaystyle \frac{1}{3}=\frac{k}{6}$$

$$\Rightarrow 3k=6$$

$$\Rightarrow k=2$$

Taking,
$$\displaystyle \frac{k}{6}=\frac{-k}{-(k+4)}$$

$$\displaystyle \Rightarrow \frac{1}{6}=\frac{1}{(k+4)}$$

$$\Rightarrow k+4=6 \Rightarrow k=2$$

Thus, $$k=2$$ is the correct answer