a) If $x_i = x_{i+1}$ for some $i$ (assuming $x_{n+1} = x_1$), then by the given identity all $x_i$ will be equal, a contradiction. Thus $x_1 \ne x_2$ and
$$x_1 - x_2 = k \frac{x_2 - x_3}{x_2 x_3}.$$
Analogously
$$x_1 - x_2 = k \frac{x_2 - x_3}{x_2 x_3} = k^2 \frac{x_3 - x_4}{(x_2 x_3)(x_3 x_4)} = \dots = k^n \frac{x_1 - x_2}{(x_2 x_3)(x_3 x_4) \dots (x_1 x_2)}.$$
Since $x_1 \ne x_2$ we get
$$x_1 x_2 \dots x_n = \pm \sqrt{k^n} = \pm k^{\frac{n-1}{2}} \sqrt{k}.$$
If among these two values, positive or negative, is obtained, then the other one will be also obtained by changing the sign of all $x_i$ since $n$ is odd.

b) From the above result, as $n$ is odd, we conclude that $k$ is a perfect square, so $k \ge 4$. For $k=4$ let $n=2019$ and $x_{3j} = 4$, $x_{3j-1} = 1$, $x_{3j-2} = -2$ for $j=1,2,\dots,673$. So, the required least value is $k=4$.