Lower bound on probability of getting two close points in a sample of $n$ points
Here's a counter-example showing your desired bound is not possible, unless I am mistaken.  It's a simple variant of the example in Roei's comment. 

Fix any $n$ and $N\ge 4n$.  Take $D$ to contain $N/2$ points that are all the same (or all within distance 1 from each other), and $N/2$ points that are all widely separated (at distance at least 1 from every other point in $D$).  This can be done even with $D\subset\mathbb{R}^1$.

The probability that two uniformly random points in $D$ are within distance 1 from each other is $1/4$.  In your notation, $p=1/4$ for $B=1$.

The probability that $n$ uniformly random points in $D$ don't contain some pair within distance 1 from each other is (at least) the probability that all $n$ points are in the widely separated set and distinct, which is at least $[(N/2-n)/N]^n \ge 1/4^n$.  

So, in your notation $1-P \ge 1/4^n \ge (3/4)^{5n} = (1-p)^{5n}$.

So, $P \le 1 - (1-p)^{5n}$.  So your desired upper bound $P \ge 1-(1-p)^{\omega(n)}$ does not hold.

(If you generalize the example above for arbitrary (small) $p&gt;0$, I think you get something like $P \le 1- (1-\sqrt p)^n \approx 1-(1-p)^{n/\sqrt p}$, which is not a counterexample if $p\rightarrow 0$ as $n\rightarrow \infty$.)