Total number of ways to place letters in envelopes = $\large n!$
Total number of derangements = $\large !n$ $=$$\Large \frac{n!}{e}$
Probability that atleast one of the letters arrive at its destination = $1 -$$\Large \frac{\frac{n!}{e}}{n!}$
= $1-$$\Large \frac{1}{e}$
Hence D is correct.
Alternate method,
Probability that letter will go in its envelope = $\Large \left ( \frac{1}{n} \right )$
$\mu (mean) = np = n \times \left ( \frac{1}{n} \right ) = 1$
$P(atleast \ 1) = 1 - P(No \ letter \ reaches \ its \ desired \ address)$
$P(r)$$\Large = \frac{m^r \times e^{-m}}{r!}$
$P(No \ letter \ reaches \ its \ desired \ address) = $$\Large \frac{m^0 \times e^{-1}}{0!}$ = $\large e^{-1}$
$P(atleast \ 1) = 1 - e^{-1}$
