Step 1: Reframe the sum using the skewness of the Poisson distribution.
$L$ is $P(N \le n)$ for $N \sim Poisson(n)$, the probability that the count is at most equal to its own mean $n$. The skewness of a Poisson($n$) distribution is $1/\sqrt{n}$, which shrinks to 0 as $n$ grows, so the distribution becomes more and more symmetric about its mean for large $n$.
Step 2: Use symmetry directly instead of invoking the CLT limit formula.
A distribution that is (in the limit) perfectly symmetric about its mean $n$ splits its total probability mass of 1 into two equal halves on either side of the mean: $P(N < n) \to P(N > n)$ as $n \to \infty$.
Since $P(N < n) + P(N = n) + P(N > n) = 1$ and the individual term $P(N=n) = \frac{e^{-n}n^{n}}{n!}$ shrinks to 0 as $n \to \infty$ (by Stirling's approximation, it behaves like $\frac{1}{\sqrt{2\pi n}} \to 0$), the two tails must each approach $1/2$.
Step 3: Match this to $L$.
$L = P(N \le n) = P(N < n) + P(N=n)$. Since $P(N<n) \to 1/2$ and $P(N=n) \to 0$, $$L \to \frac{1}{2} + 0 = 0.5$$
Step 4: Sanity-check with the boundary options.
$L=1.0$ would require the sum to go on forever past $k=n$, capturing the whole distribution, which is not what the given sum does. $L=0$ contradicts the fact that roughly half the mass sits at or below the mean by symmetry. $e^{-1}$ is just the isolated $k=0$ term, not the cumulative sum.
Conclusion:
This confirms $L = 0.5$, option (A), matching a well known asymptotic result for the Poisson CDF evaluated at its own mean.$$\boxed{L = 0.5}$$