Find P if, $8=3+\frac{1}{4}(3+p)+\frac{1}{4^2}(3+2p)+\frac{1}{4^3}(3+3p)+\dots <!--\; \dots \; -->\mathrm{\infty <!--\; \infty \; -->}$ |

The correct answer is 9.

To find the value of $p$, we analyze the given infinite series:
$$8=3+\frac{1}{4}(3+p)+\frac{1}{4^2}(3+2p)+\frac{1}{4^3}(3+3p)+\dots$$

This expression is an Arithmetico-Geometric Progression (AGP). Let the sum of this infinite series be denoted by $S$:
$$S=3+\frac{3+p}{4}+\frac{3+2p}{4^2}+\frac{3+3p}{4^3}+\dots \infty$$             — (Equation 1)

The common ratio of the geometric part of this progression is $\frac{1}{4}$. To solve the AGP, we multiply the entire series by the common ratio:
$$\frac{1}{4}S=\frac{3}{4}+\frac{3+p}{4^2}+\frac{3+2p}{4^3}+\dots \infty$$             — (Equation 2)

Now, we subtract Equation 2 from Equation 1 by aligning the terms with the same denominators:
$$S-\frac{1}{4}S=3+(\frac{3+p}{4}-\frac{3}{4})+(\frac{3+2p}{4^2}-\frac{3+p}{4^2})+(\frac{3+3p}{4^3}-\frac{3+2p}{4^3})+\dots$$

Simplifying the subtracted terms, we get:
$$\frac{3}{4}S=3+\frac{p}{4}+\frac{p}{4^2}+\frac{p}{4^3}+\dots$$

We can factor out $p$ from the terms starting from the second term:
$$\frac{3}{4}S=3+p(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\dots )$$

The term inside the parentheses is an infinite geometric series with the first term $a=\frac{1}{4}$ and the common ratio $r=\frac{1}{4}$. The sum of an infinite geometric series is given by the formula:
$$S_{\infty }=\frac{a}{1-r}$$

Calculating the sum of the geometric series:
$$\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\dots =\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$$

Substituting this sum back into our equation:
$$\frac{3}{4}S=3+\frac{p}{3}$$

Given that the total sum of the series is $S=8$, we substitute this value to solve for $p$:
$$\frac{3}{4}(8)=3+\frac{p}{3}$$

Simplify the left side:
$$6=3+\frac{p}{3}$$

Subtract 3 from both sides:
$$3=\frac{p}{3}$$

Multiply both sides by 3:
$$p=9$$