I suspect the confusion may be in what you mean by "step" in the phrase "$N$ steps ahead".
If you mean the one-second time step of your discrete time Markov process, then what you're calculating is exactly correct — with a $0.98$ self-loop transition probability, the process has a less than $0.5$ probability of moving anywhere within $N < 35$ steps.
If, however, by "step" you mean a transition in which the process does not stay in the same state, then you need to eliminate the self-loop transitions from your matrix (since you don't count them as steps) by setting their probability to $0$ and normalizing the remaining transition probabilities for each state so that they again sum to $1$.
Of course, what you lose by doing so is the information about the residence time in each state. If you wanted to know where the process will be after $N$ non-self-loop steps and how long it'll take to get there, that would require a rather more complicated calculation, and the resulting discrete phase-type distribution would, in general, be unlikely to have a simple form (although it might be possible to approximate it with some nicer distributions).
In the comments, you write that your transition matrix is (approximately) $$P=\pmatrix{0.99&0.0208&0&0\\0.0019&0.9792&0.0104&0\\0&0&0.9896&0.0108\\0‌​.0028&0&0&0.9907}$$ and that "second option it is what I'm looking for", i.e. that you want to calculate the probability distribution after $N$ steps, a step denoting a transition between two distinct states. If so, what you need to do is to calculate another stochastic matrix $P'$ which describes the same process, but which does not include the self-loop transitions (i.e. whose diagonal entries are zero).
To obtain $P'$ from $P$, simply zero out the diagonal entries and re-normalize each row so that the matrix stays stochastic (i.e. so that each row sums to $1$). Since most of the rows in $P$ only have one non-zero off-diagonal element to begin with, that element becomes $1$, leaving row 2 as the only non-trivial case: $$P' = \pmatrix{0&1&0&0\\0.1545&0&0.8455&0\\0&0&0&1\\1&0&0&0}$$ From this, you can calculate the state distribution $\pi_n = \pi P'^n$ after $n$ steps, given a starting distribution $\pi$.
However, note that, whereas the original transition matrix $P$ was ergodic, $P'$ isn't — in particular, a process in an even-numbered state always ends up in an odd-numbered state after one step and vice versa, so all states have an even period. This lack of ergodicity doesn't really affect the calculation of $\pi_n$ as such, but it does mean e.g. that $\pi_n$ won't generally converge to a stationary distribution as $n \to \infty$.
Best Answer
For starters, you have the wrong transition matrix. You want $P$ to be such that
$$P{1\brack 0}={\frac{1}{3}\brack \frac{2}{3}}\;,$$
meaning that if you start with something in-state $A$, after one transition it’s still in state $A$ with probability $\frac13$ and in-state $B$ with probability $\frac23$. Similarly, you want
$$P{0\brack 1}={4/5\brack 1/5}\;,$$
meaning that if you start with something in state $B$, after one transition it’s still in state $B$ with probability $\frac15$ and has moved over to state $A$ with probability $\frac45$. This means that you want
$$P=\begin{bmatrix}1/3&4/5\\2/3&1/5\end{bmatrix}\;.$$
For $n\in\Bbb N$ let $B_n$ be the $2\times 1$ matrix whose top entry is the probability that $X_n$ is $A$, and whose bottom entry is the probability that $X_n$ is $B$. You’re told that $$B_0={1/2\brack1/2}\;:$$ the process is equally likely to start in each of the states, so initially it’s in each state with probability $\frac12$. You also know that $B_{n+1}=PB_n$ for each $n\in\Bbb N$: that’s simply how the Markov process works. Thus,
$$B_1=\begin{bmatrix}1/3&4/5\\2/3&1/5\end{bmatrix}B_0=\begin{bmatrix}1/3&4/5\\2/3&1/5\end{bmatrix}\begin{bmatrix}1/2\\1/2\end{bmatrix}=\begin{bmatrix}17/30\\13/30\end{bmatrix}\;,$$
where I’ll leave the conclusion of the calculation to you.
Once you have this, you need to find the probability that $X_1=X_2$. That’s the probability that
$$(X_1=A\text{ and }X_2=A)\quad\text{or}\quad(X_1=B\text{ and }X_2=B)\;;$$
can you find that? Remember what the transition probabilities mean.