Exact probability for the distance after $t$ steps
we can use a method from G. A. Whitmore & V. Seshadri to derive the first passage time of a Wiener process (described in Deriving Inverse Gaussian as First Passage Time of Wiener Process) to compute an exact distribution.
Let $k(t)$ be the position of the drunk man after $t$ steps, $k(0) = 1$, the absorbing boundary is at $k=0$, the man takes steps forward with probabability $p$ and backwards with probabability $q=1-p$, and for simplicity we consider even numbers of steps $t$.
For a given path that the drunk man takes he took $x=\frac{t+k(t)}{2}$ steps forward and $y=\frac{t-k(t)}{2}$ steps backwards, and the particular probability for that path is $p^xq^y$.
The number of paths that lead to position $k(t)$ is equal to ${{t}\choose{x}} - {{t}\choose{x+1}}$ if $x<t$ or $1$ if $x=1$. This can be argued based on a reflection principle. One can imagine a random walk without absorbing boundary at zero, and consider the paths that ended up at zero or below.
For every such path that ended up below zero, we can imagine a reflected path that ended up above zero.
Such reflected paths are all the possible paths that ended up above zero, but hit the zero in between time $0$ and $t$, thus out of the ${{t}\choose{x}}$ paths that end up in $x$ if there is no absorbing boundary, ${{t}\choose{x+1}}$ are paths that can be reflected.
Thus, the probability is
$$P(K = k|t) = \left[{{t}\choose{x}} - {{t}\choose{x+1}}\right] p^xq^y$$
where $x=\frac{t+k}{2}$ steps and $y=\frac{t-k}{2}$
If we compute the integral you get
$$P(K >0 |t) = S(t+1,2t,p) - \frac{q}{p} S(t+2,2t,p)$$
where $S(x,t,p)$ is the survival function of the binomial distribution.
In the limit $t \to \infty$ we get $$\lim_{t\to\infty} P(K >0 |t) = 1-\frac{q}{p} = 2 - \frac{1}{p}$$ this is the same as the earlier computed $1-\frac{1}{2p} + \frac{\sqrt{1-4p(1-p)}}{2p}$ where the term in the root could have been simplified as $(1-2p)^2$.
