The correct answer is: $\boxed{\text{D}. F\left( s \right) = \frac{1}{{1 – {e^{ – sT}}}}}$
The Laplace transform of a periodic function $f(t)$ with period $T$ is given by:
$$F(s) = \sum_{n=-\infty}^{\infty} \frac{f(nT)}{s+nT}$$
In this case, $f(t)$ is a square wave with period $T$, so:
$$f(t) = \begin{cases} 1 & \text{for } 0 \leq t \leq T \\ 0 & \text{for } t > T \end{cases}$$
Substituting this into the Laplace transform formula, we get:
$$F(s) = \sum_{n=-\infty}^{\infty} \frac{1}{s+nT} = \frac{1}{1-e^{-sT}}$$
Therefore, the Laplace transform of the causal periodic square wave of period $T$ is $\frac{1}{1-e^{-sT}}$.
Option A is incorrect because it does not include the factor of $1-e^{-sT}$. Option B is incorrect because it does not include the factor of $s$. Option C is incorrect because it does not include the factor of $1-e^{-sT}$.