it's AGP (Arithmetico-Geometric Progression)

it becomes

3( 1 + $\frac{1}{4} + \frac{1}{4^{2}} .......................$ ) + p ( $\frac{1}{4} + \frac{2}{4^{2}} .......................$ )

now let the 2nd series be equal to x

therefore,

x= ( $\frac{1}{4} + \frac{2}{4^{2}} .......................$ )

$\frac{x}{4}$ = ( $\frac{1}{4^{2}} + \frac{2}{4^{3}} .......................$ )

then

x - $\frac{x}{4}$ = ( $\frac{1}{4} + \frac{1}{4^{2}} .......................$ )

therefore x = 4/3( $\frac{1}{4} + \frac{1}{4^{2}} .......................$ )

and we know the sum of an G.P

then finally it comes

4 + $\frac{4P}{9}$ = 8

thus P=9

There is an outstanding expression that rules this kind of series,

$S=\dfrac{a}{1-r}+\dfrac{dr}{{(1-r)^2}} \quad \quad |r|<1$

To the sequences that look like,

$a , (a+d) r , (a+2d) r^2 , (a+3d)r^3, \ldots , \left[ a + (n-1) d \right] r^{n-1},\ldots$

In this particular case $a=3$ , $d=p$ and $r=\frac 14$ ,

$\dfrac 3{1-\dfrac 14} + \dfrac {\dfrac p4}{\left( 1- \dfrac14 \right)^2} =8 \Leftrightarrow 4+\dfrac {4p}9 =8 \Leftrightarrow 36+4p=72 \Leftrightarrow \boxed{p=9}$ $\square$

References

I also did the same way