A Poisson process with a fixed maximum number of counts? For a Poisson process, we have the pmf of arrival in (0,t] as:
$$p(n)=\frac{(\lambda t)^n \exp(-\lambda t)}{n!}$$
where $\lambda$ is the arrival rate and the PDF of the inter-arrival times is given as:
$$f(t)=\lambda \exp(-\lambda t)$$
What if I assume that the maximum number of counts that is possible is $N$, I mean in case of a call scenario the maximum calls can never exceed beyond say $N$. For such a scenario:
$\sum_{n=1}^{500}p(n)=1$ and $\sum_{n=501}^{600}p(n)=0$. Can w incorporate such a scenario in the Poisson processes?

If I understand your question correctly, then I believe such a distribution is possible. Your original probability mass function was defined by
$$P(n\text{ arrivals in }(0,t])=p(n)=\frac{(\lambda t)^n e^{-\lambda t}}{n!}$$
If you wish to limit the “number of arrivals” to $N$, then the new probability mass function will be given by the following, for $n\le N$:
$$\begin{align}P(n\text{ arrivals in }(0,t] \space | \text{ at most N arrivals})&=\frac{P(n\text{ arrivals in }(0,t])}{P(\text{at most N arrivals})}\\
&= \frac{p(n)}{\sum_{k=0}^N p(k)}\\
 \end{align}$$
Which should give the distribution you are looking for.
