The correct answer is $\frac{F(s)}{1-e^{-sT}}$.
The Laplace transform of a function $f(t)$ is defined as
$$L[f(t)] = \int_0^\infty f(t) e^{-st} dt$$
If $L[f(t)] = F(s)$, then the Laplace transform of $f(t-T)$ is
$$L[f(t-T)] = \int_0^\infty f(t-T) e^{-st} dt = \int_T^\infty f(t) e^{-(s-T)t} dt = \int_0^\infty f(t) e^{-st} dt e^{-Ts} = F(s) e^{-Ts}$$
Therefore, the Laplace transform of $f(t-T)$ is $\frac{F(s)}{1-e^{-sT}}$.
Option A is incorrect because it does not take into account the time delay of $T$. Option B is incorrect because it takes the time delay into account, but it does so incorrectly. Option C is incorrect because it is the Laplace transform of a shifted exponential function, not a delayed exponential function. Option D is correct because it is the Laplace transform of a delayed exponential function.