P.S-: I am editing my Answer as Previous answer had some error.
$\Rightarrow$ Let $U_{n}$ be the Event that a Person is using the product in the $nth$ month.
Given$P\left ( U_{1} \right )=0.25$
$\Rightarrow$ Probablity that if person is using the product in the $nth$ month,he will use it in next month$\Rightarrow$
$P \left ( \left( U_{n+1} \right )|P\left ( U_{n} \right ) \right )=$0.80
$\Rightarrow$ Probablity that if person is not using the product in the $nth$ month,he will use it in next month$\Rightarrow$
$P \left ( \left( U_{n+1} \right )|P\left ( U_{n} \right ){}' \right )=$0.30
Let us first find the probablity that the person uses the product in next month
$P\left ( U_{n+1} \right )=P\left (U_{n+1} | U_{n} \right )*P\left ( U_{n} \right )+P\left (U_{n+1} | U_{n}{}' \right )*P\left ( U_{n}{}' \right )$
$P\left ( U_{n+1} \right )=0.80*P\left ( U_{n} \right )+0.30*P\left ( U_{n}{}' \right )$
$P\left ( U_{n+1} \right )=0.80*P\left ( U_{n} \right )+0.30*\left ( 1- P\left ( U_{n} \right )\right )$
$P\left ( U_{n+1} \right )=0.5* P\left ( U_{n} \right )+0.3$
Solving the recurrence relation
Let $P_{n+1}=P\left ( U_{n+1} \right )$, with $P\left ( U_{1} \right )=0.25$
So our reccurence will be simplified to
$P_{n+1}=0.5*P_{n}+0.3$
our roots are $r-0.5=0\Rightarrow 0.5$
Solving for $P_{n} $,
$P_{n}=\alpha * root^{n}+\beta \Rightarrow P_{n}=\alpha * 2^{-n}+\beta$
$P_{n+1}=\alpha * 2^{-\left ( n+1 \right )}+\beta$$\cdots \cdots \cdots \cdots eqtn 1$
Also ,$P_{n+1}=0.5*P_{n}+0.3$ $
so
$P_{n+1}=0.5 *\alpha * 2^{-n}+\beta+0.3$
=$1/2 \left ( \alpha * 2^{-n}+\beta \right )+0.3$
=$\left ( \alpha * 2^{-\left ( n+1 \right )}+\beta /2 \right )+0.3$$\cdots \cdots \cdots \cdots eqtn 2$
From eqtn 1 and 2,
$\left ( \alpha * 2^{-\left ( n+1 \right )}+\beta /2 \right )+0.3=\alpha * 2^{-\left ( n+1 \right )}+\beta$
$\Rightarrow \beta=6/10$
Given that $P\left ( U_{1}\right )=25$
We have to find $\alpha$
we have $P_{n}=P\left ( U_{n}\right )=\alpha * 2^{-n}+\beta$
take n=1$\Rightarrow P_{1}=P\left ( U_{1}\right )=\alpha * 2^{-1}+0.6=25$
$\Rightarrow \alpha * 0.5^{1}+0.6=25 \Rightarrow \alpha \approx 24$
we have
$P_{n+1}=\alpha * 2^{-\left ( n+1 \right )}+\beta$
$P_{n+1}=24* 2^{-\left ( n+1 \right )}+0.6$
Taking $n$ as $\infty$(As we are not given a limited month tenure for which it will last ,it is asking for "long time ")
$P_{n+1}=0.6$
Thus Answer =60%