The correct answer is $\boxed{\text{B. }A(s) = \frac{1}{1 – \exp(-Ts)}}$.
The Laplace transform of a function $f(t)$ is defined as
$$Lf(t): s = \int_0^\infty f(t) e^{-st} dt$$
If $f(t)$ is periodic with period $T$, then the Laplace transform of $f(t)$ can be written as
$$Lf(t): s = A(s) \int_0^T f(t) e^{-st} dt$$
where $A(s)$ is the Laplace transform of the first period of $f(t)$.
To find $A(s)$, we can use the following formula:
$$Lf(t): s = \frac{1}{1 – \exp(-sT)} \int_0^T f(t) e^{-st} dt$$
Substituting this into the definition of $Lf(t): s$, we get
$$A(s) = \frac{1}{1 – \exp(-sT)}$$
Therefore, the correct answer is $\boxed{\text{B. }A(s) = \frac{1}{1 – \exp(-Ts)}}$.
The other options are incorrect because they do not take into account the periodicity of $f(t)$.