Since we are looking at an infinite number of trials, after removing a finite prefix of trials, we still have an infinite number of trials. Therefore, we can simply ignore this prefix for the probability of $E$ (even the conditions for $E$ would already be fulfilled by the prefix) and therefore $P(E) = P(E \mid B)$ for any event $B$ that restricts the outcomes of the trials only for a finite prefix.
This hopefully answers your question, but the partial solution you gave above seems rather complicated to me. Thankfully, I was able to find a pdf version of the book via Google. There, the author gives a rather complicated probability for $P(E)$ (not included in your question). That surprised me, because intuitively I thought, $P(E) = 1$.
Thinking about that in more detail, I am now convinced that, indeed, $P(E) = 1$: Let $A$ be the event, that we consecutively have $n$ successes and $m$ failures in $n + m$ trials. Clearly, $P(A) ≠ 0$, because $P(A) = p^n ⋅ q^m$ (assuming $0 < p < 1$).
Now let us iterate these $n + m$ trials and let $E'$ be the event that event $A$ occurs in the first block of $n + m$ trials, or in the second block of $n + m$ trials, or in the third block, etc. Since $P(A) ≠ 0$ and therefore $P(\overline{A}) ≠ 1$ it is easy to see that $P(\overline{E'}) = 0$, since we are looking at an infinite number of trials. Therefore we have $P(E') = 1$.
Since $E' ⊆ E$, it follows that also $P(E) = 1$.
For short: In an infinite sequence of trials you are able to find any finite pattern with probability $1$.
Therefore I suspect that there is a mistake in the solution in the book – unless I made a mistake myself or misunderstood the problem.
Best Answer
The count of successes among a specified amount of Bernoulli trials each with an independent and identically distributed success rate, is a Binomially Distributed random variable.
Order of the sequence is not one of these criteria.
However, when drawing simultaneously, the results may not be independent. If they are not, you may instead have a hypergeometric distribution.