A simplified version of the given program can be written as:

float f(float x, int y) {
float p=1, s=1;
int i;
for (i=1; i<y; i++) {
p = p * (x/i);
s = s + p;
}
return s;
}

We can take $p=1,s=1$ initialization outside of for loop because there is no condition checking in for loop involving $p,s$ .

$$\begin{array}{|l|l|l|} \hline \text{$i$} & \text{$p= p $*$ (x/i)$} & \text{$s=s+p$}\\\hline \text{1} & \text{$x$} & \text{$1+x$} \\\hline \text{2} & \text{$\frac{x^{2}}{2}$} & \text{$1+x+\frac{x^{2}}{2}$}\\\hline \text{3} & \text{$\frac{x^{3}}{6}$} & \text{$1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}$} \\\hline \text{4} & \text{$\frac{x^{4}}{24}$} & \text{$1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\frac{x^{4}}{24}$}\\\hline \text{n} & \text{$\frac{x^{n}}{n!}$} & \text{$e^x$} \\\hline \end{array}$$

$$\begin{array}{|l|} \hline \text{$e^x = \displaystyle{\sum_{n=0}^{\infty} \frac{x^n}{n!}} = 1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6} +\frac{x^{4}}{24}+ \ldots+\frac{x^{n}}{n!}$} \\\hline \end{array}$$

Hence, option **B** is answer.