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.

We are given the following infinite series equation:
$8=3+\frac{1}{4}(3+p)+\frac{1}{4^2}(3+2p)+\frac{1}{4^3}(3+3p)+...\infty$

This is an Arithmetico-Geometric Progression (AGP). Let us denote the sum of the infinite series on the right side as S:
$S=3+\frac{3+p}{4}+\frac{3+2p}{4^2}+\frac{3+3p}{4^3}+...$

The general formula for the sum of an infinite Arithmetico-Geometric Progression (A.G.P.) of the form:
$S=a+(a+d)r+(a+2d)r^2+...$
where $|r|<1$ is given by:
$S=\frac{a}{1-r}+\frac{dr}{{(1-r)}^2}$

Comparing this with our series:
The first term of the arithmetic part, $a=3$
The common difference of the arithmetic part, $d=p$
The common ratio of the geometric part, $r=\frac{1}{4}$

Substituting these values into the sum formula:
$8=\frac{3}{1-\frac{1}{4}}+\frac{p\cdot \frac{1}{4}}{{(1-\frac{1}{4})}^2}$

Simplify the term $1-\frac{1}{4}=\frac{3}{4}$:
$8=\frac{3}{\left(\frac{3}{4}\right)}+\frac{\left(\frac{p}{4}\right)}{{\left(\frac{3}{4}\right)}^2}$

Calculate each part:
First part:
$\frac{3}{\left(\frac{3}{4}\right)}=3\cdot \frac{4}{3}=4$
Second part:
$\frac{\left(\frac{p}{4}\right)}{\left(\frac{9}{16}\right)}=\frac{p}{4}\cdot \frac{16}{9}=\frac{4p}{9}$

Now, substitute these simplified terms back into our equation:
$8=4+\frac{4p}{9}$

Subtract 4 from both sides:
$8-4=\frac{4p}{9}$
$4=\frac{4p}{9}$

Multiply both sides by 9:
$36=4p$

Divide both sides by 4:
$p=9$