Probability of going into an absorbing state

Let $L_i$ be the likelihood that you terminate at node $0$ given that you start at node $i$.

Clearly, $L_0 = 1,\ L_3 = 0$. In your case $p = p_{i,i-1} = 0.3 = 1 - p_{i-1,i}$, in general it can be shown that $$L_1 = pL_0 + (1-p)L_2\text{, and }L_2 = pL_1 + (1-p)L_3$$

It's clear then that $L_1 = (1)p+(1-p)L_2 = p + (1-p)(pL_1 + (1-p)(0))$, so $$L_1 = p + (1-p)pL_1 \implies L_1 = \frac{p}{1 - (1-p)p}$$ $$L_2 = pL_1 = \frac{p^2}{1 - (1-p)p}$$

Hint:

Another approach to solving this problem is to see what the sum of the amounts that are absorbed at each step are.

Let $S$ be the initial state and $$ M=\begin{bmatrix} 0&0&0&0\\ 0.3&0&0.7&0\\ 0&0.3&0&0.7\\ 0&0&0&0 \end{bmatrix} $$ Then the absorbent terms (the left and right elements) of $SM^k$ are the amounts that gets absorbed at step $k$. Thus, the absorbent terms of $$ S\left(M+M^2+M^3+M^4+\cdots\right)=SM(I-M)^{-1} $$ tell how much has been absorbed in total.

The probability that we get to state zero immediately is $0.3$. The next possibility is that we get to state two then we get back to state one and then to state zero, the probability of which event is $0.7\cdot0,3\cdot0.3=0.7\cdot0.3^2$. The probability of the next possibility is $0.7\cdot0.3\cdot0.7\cdot0.3\cdot0.3=0.7^2\cdot0.3^3$, and so on.

The probability that we get to state zero once in the future is then $$\sum_{i=0}^{\infty} 0.7^{\ i}0.3^{\ i+1}=0.3\sum_{i=0}^{\infty} 0.21^{\ i}=0.3\frac{1}{1-0.21}.$$