It will turn out that the number of letters processed has Poisson distribution with parameter $\lambda(1-0.04)$. There are better conceptual ways of seeing this, but instead we do a somewhat unpleasant computation.
The Post Office receives $n$ letters with probability $e^{-\lambda}\dfrac{\lambda^n}{n!}$. Given that it receives $n$ letters, the number of letters processed has Binomial distribution. Specifically, the (conditional) probability that it processes $k$ letters is $\binom{n}{k}p^kq^{n-k}$. (In our case, $p=0.96$ and $q=1-p=0.04$.)
So the probability that the Post Office processes exactly $k$ letters is given by $\sum_{n=k}^\infty e^{-\lambda}\frac{\lambda^n}{n!}\binom{n}{k}p^kq^{n-k}.$ We are interested specifically in the sum for $k=5$. But we continue with the general analysis.
Use $\dbinom{n}{k}=\dfrac{n!}{k!(n-k)!}$ and some algebra to rewrite the above expression as $e^{-\lambda}\frac{(p\lambda)^k}{k!} \sum_{n=k}^\infty\frac{1}{(n-k)!} (\lambda q)^{n-k}.$ The $\sum_{n=k}^\infty$ part, perhaps after putting $j=n-k$, can be seen to be the power series expansion of $e^x$, with $x=\lambda q$. But $e^{-\lambda}e^{\lambda q}=e^{-\lambda p}$, so our expression simplifies to $e^{-\lambda p}\frac{(\lambda p)^k}{k!}.$ It follows that the number of letters processed has Poisson distribution, parameter $\lambda p$. In our case, $\lambda p=4.8$. Now finding the probability that $k=5$ is easy.