The correct answer is $\boxed{\text{D. Undefined}}$.
The Laplace transform of a function $f(t)$ is defined as
$$F(s) = \int_0^\infty f(t) e^{-st} dt$$
If $f(t)$ is a piecewise continuous function on $[0, \infty)$, then the Laplace transform exists for all $s$ such that $\Re(s) > 0$.
In this case, we are given that
$$F(s) = L\left| {f\left( t \right)} \right| = {K \over {\left( {s + 1} \right)\left( {{s^2} + 4} \right)}}$$
We can see that $F(s)$ is a rational function with no poles in the right half-plane. Therefore, $F(s)$ is analytic in the right half-plane and can be extended to an entire function.
The Laplace transform of a function is unique if the function is zero for $t \ge T$ for some $T$. In this case, we are not given any information about $f(t)$ for $t \ge T$. Therefore, we cannot conclude that $f(t)$ is zero for $t \ge T$.
As a result, we cannot conclude that $\lim_{t \to \infty} f(t)$ exists. Therefore, the answer is $\boxed{\text{D. Undefined}}$.
Here is a brief explanation of each option:
- Option A: $${K \over 4}$$
This is the value of $F(s)$ at $s = 0$. However, we cannot conclude that $\lim_{t \to \infty} f(t) = {K \over 4}$ because $F(s)$ is not necessarily the Laplace transform of $f(t)$.
- Option B: Zero
This is the value of $f(t)$ at $t = 0$. However, we cannot conclude that $\lim_{t \to \infty} f(t) = 0$ because $f(t)$ is not necessarily continuous at $t = 0$.
- Option C: Infinite
This is the value of $f(t)$ at $t = \infty$. However, we cannot conclude that $\lim_{t \to \infty} f(t) = \infty$ because $f(t)$ is not necessarily defined at $t = \infty$.
- Option D: Undefined
This is the correct answer because we cannot conclude that $\lim_{t \to \infty} f(t)$ exists.