for determining the probabilities, it has been written on the book that:
$$p_n=\frac{\frac{d^n}{ds^n}P(s)|_{s=0}}{n!};\ldots(A)$$
But if i set $s=0$ then $p_n$ becomes $0$.
$$p_1=\frac{\frac{d}{ds}P(s)|_{s=0}}{n!}$$ $$\Rightarrow p_1=\frac{\frac{d}{ds}\mathbb E[s^X]|_{s=0}}{n!}$$ $$\Rightarrow p_1=\frac{\mathbb E[Xs^{(X-1)}]|_{s=0}}{n!}$$ $$\Rightarrow p_1=\frac{\mathbb E[X0^{(X-1)}]}{n!}$$ $$\Rightarrow p_1=\frac{\mathbb E[0]}{n!}$$ $$\Rightarrow p_1=\frac{0}{n!}=0$$
Can you please prove the relation in equation (A) and check it for Binomial Distribution where the generating function of the Binomial Distribution is $P(s)=(1-p+ps)^n$, where $p$ denotes probability of success.
We try to show that all that is involved in the probability generating function is familiar calculus ideas. We have a random variable $X$ that takes on non-negative integer values only. Then $$P(s)=E(s^X)=p_0+p_1s+p_2s^2+p_3s^3+\cdots.$$ We could write the first term as $p_0s^0$, on the understanding that $s^0$ is identically $1$. But $p_0$ is cleaner.
The $p_i$ are non-negative and less than $1$. So the power series above has radius of convergence at least equal to $1$.
The $p_k$ obviously determine $P(s)$. We show that the $p_k$ can be uniquely recovered from $P(s)$. This amounts to showing that we can recover the $p_k$ by a derivatives calculation.
In the interior of the interval of convergence, we can freely differentiate term by term. Now the first formula of your post is the familiar assertion that the coefficient of $s^n$ in the Maclaurin series of a function $P(s)$ is $\frac{1}{n!} P^{(n)}(0)$. We can get at it mechanically by differentiating the series on the left term by term $n$ times and setting $s=0$, When we differentiate $n$ times, all the terms before the $s^n$ term die. The $n$-th derivative of $s^n$ is $n!$, and the derivatives of all the terms that involve powers of $s$ higher than $n$ still have an $s$ in them, so die when we set $s=0$.
To illustrate with the binomial distribution, differentiate $(1-p+ps)^n$ $k$ times, and set $s=0$. Differentiating $k$ times gets us a $p^k$ from the Chain Rule applied $k$ times. Then we get $(n)(n-1)\dots(n-k+1)$, which when divided by $k!$ is just $\binom{n}{k}$. And after we have differentiated $k$ times, there remains $(1-p+ps)^{n-k}$, which becomes $(1-p)^{n-k}$ when we set $s=0$. Thus we have, using the formula at the beginning of your post, recovered the probabilities $p_k=\binom{n}{k}p^k(1-p)^{n-k}$.