We should be able to get a quick answer for the case when Player A has infinite capital. The number of steps until B's stake is gone will be a negative binomial random variable. The expected value is given by $$E[T]={b(1-p) \over p}$$ For the example you give we get $$E[T]={5(0.7) \over 0.3}= 11.6667$$
Now in reality Player A will sometimes lose before B runs out, so I am thinking the conditional mean is lower.
Follow-Up: Using Andre's definition of $N$ and his solution, the mean number of steps before B runs out, conditioned on B losing, is $$E[T] = 11.1584$$
We provide an answer and link it with already given answers which might help to see connections.
Some observations:
We can reduce the problem to a combinatorial one by counting all paths starting from $(0,k)$ to $(m-1,n-1)$ using steps $(1,1)$ and $(1,-1)$ which do not reach the lines $y=0$ and $y=n$.
The starting point represents the $k$ coins of the gambler he has right at the beginning. Winning a round increases his coins by one which is represented by a step $(1,1)$ and loosing a round means also going in $x$-direction by one, but decreasing $y$ by one, so we make a step $(1,-1)$.
Each valid path from $(0,k)$ to $(m-1,n-1)$ has length $m-1$ and is realised with probability $\frac{1}{2^{m-1}}$. In order to reach $(m,n)$ this can only be done in one step from $(m-1,n-1)$ to $(m,n)$ with probability $\frac{1}{2}$, so that the number of valid paths from $(0,k)$ to $(m-1,n-1)$ has finally to be divided by $2^m$ to find the wanted probability.
We start with an example confirming the approach of @GCab.
Example (part 1): k=2, m=14, n=6
We count the number of valid paths from $(0,2)$ to $(14,6)$, which is the number of lattice paths from $(0,2)$ to $(13,5)$ which do not touch the lines $y=0$ and $y=6$, followed by an $m$-th step from $(13,5)$ to $(14,6)$.
We see in accordance with the table presented by @GCab that we have $\color{red}{364}$ valid paths which is marked red in the graphic below.
We can normalise the situation by shifting $(0,k)$ to $(0,0)$ and consider the equivalent problem to count the number of lattice paths from $(0,0)$ to $(m-1,n-1-k)$ using steps $(1,1)$ and $(1,-1)$ without reaching the boundaries $y=n-k$ and $y=-k$. We denote this number of valid paths by
\begin{align*}
L_{m-1,n-1-k;n-k,k}
\end{align*}
Formula:
The formula above in the form $L_{m,n;r,s}$ counting valid paths from $(0,0)$ to $(m,n)$ which do not reach the boundaries $y=r$ and $y=-s$ is established in this post. It can be written as
\begin{align*}
L_{m,n;r,s}&=\binom{m}{\frac{m+n}{2}}-\sum_{j\geq1}\left[\binom{m}{\frac{m+n}{2}-r+j(r+s)}
+\binom{m}{\frac{m+n}{2}+s-j(r+s)}\right]\\
&\qquad\qquad\qquad+\sum_{j\geq1}\left[\binom{m}{\frac{m+n}{2}+j(r+s)}
+\binom{m}{\frac{m+n}{2}-j(r+s)}\right]\tag{1}\\
\end{align*}
In the current situation we obtain from (1) the number of valid paths for OP's problem:
\begin{align*}
&\color{blue}{L_{m-1,n-1-k;n-k,k}}=\binom{m-1}{\frac{m+n-k}{2}-1}\\
&\quad\qquad-\sum_{j\geq1}\left[\binom{m-1}{\frac{m+n-k}{2}-1-n+k+jn}+\binom{m-1}{\frac{m+n-k}{2}-1+k-jn}\right]\\
&\quad\qquad+\sum_{j\geq1}\left[\binom{m-1}{\frac{m+n-k}{2}-1+jn}+\binom{m-1}{\frac{m+n-k}{2}-1-jn}\right]\tag{2}\\
&\quad=-\sum_{j\geq0}\binom{m-1}{\frac{m+n-k}{2}-1+k+jn}-\sum_{j\geq1}\binom{m-1}{\frac{m+n-k}{2}-1+k-jn}\\
&\quad\qquad+\sum_{j\geq0}\binom{m-1}{\frac{m+n-k}{2}-1+jn}+\sum_{j\geq1}\binom{m-1}{\frac{m+n-k}{2}-1-jn}\tag{3}\\
&\quad\,\,\color{blue}{=\sum_{j=-\infty}^{\infty}\left[\binom{m-1}{\frac{m+n-k}{2}-1+jn}-\binom{m-1}{\frac{m+n+k}{2}-1+jn}\right]}\tag{4}
\end{align*}
Comment:
In (3) we shift in the left-most series the index by one to start with $j=0$. In the third series we merge the single left-most term from (2).
In (4) we combine the two-right most series as well as the two left-most series.
The resulting probability is
\begin{align*}
\color{blue}{\frac{1}{2^m}L_{m-1,n-1-k;n-k,k}}
\end{align*}
The sums in (2) are a consequence of applying the principle of inclusion-exclusion to reflected paths. This is necessary in order to compensate double counting as indicated in the answer from @Hans.
.
Example (part 2): k=2, m=14, n=6
In order to check (2) we calculate the number of valid paths from the example above.
We obtain
\begin{align*}
\color{blue}{L_{13,3;4,2}}&=\binom{13}{8}-\left[\binom{13}{10}+\binom{13}{4}\right]+\left[\binom{13}{2}\right]\tag{3}\\
&=1\,287-(286+715)+78\\
&\,\,\color{blue}{=364}
\end{align*}
in accordance with the first part of the example. Note the number of reflected paths in the brackets in (3) are indicated in the graphic by $A_1, B_1$ and $B_2$.
Best Answer
Your approach is correct, mistake is in initial or boundary condition. $P_a=0$ means that game stops when $k=a$. According to the problem text, game stops when $k=0$.
You can save your solution if you will mean $k$ is opponent capital. Then answer is $E_{a-k}$ of your solution which is equal to $k(2a-1-k)$.