Answer
$\color{blue}{f_X(x) = p(1-p)^{x-1}, x=1,2,3,\ldots}.$
Work Step by Step
For $X=1,$
$\begin{align*}
f_X(1) &= P(X\le 1) - P(X\lt 1) \\
&= F(1) - 0 & [\text{since}\ F_X(x)\ne 0,\ \text{only for}\ x=1,2,3,\ldots] \\
&= (1-(1-p)^1) - 0\\
&= 1-1+p \\
f_X(1) &= p.
\end{align*}$
For $X=2,3,4\ldots,$
$\begin{align*}
f_X(x) &= P(X=x) \\
&= P(X\le x) - P(X\lt x),\ x=2,3,4,\ldots \\
&= P(X\le x) - P(X\le x-1) & [\text{since}\ F_X(x)\ne 0,\text{only for}\ x=1,2,3\ldots] \\
&= F(x) -F(x-1) \\
&= (1-(1-p)^x) - (1-(1-p)^{x-1}) \\
&= 1-(1-p)^x -1 + (1-p)^{x-1} \\
&= -(1-p)^x + (1-p)^{x-1} \\
&= -(1-p)(1-p)^{x-1} + (1-p)^{x-1} \\
&= (-(1-p)+1)(1-p)^{x-1} \\
&= (-1+p+1)(1-p)^{x-1} \\
f_X(x) &= p(1-p)^{x-1},\ x=2,3,\ldots.
\end{align*}$
Since $f_X(1) = p = p(1-p)^0 = p(1-p)^{1-1},$ we can then define $\color{blue}{f_X(x) = p(1-p)^x, x=1,2,3,\ldots}.$
