The probability of player $1$ being a winner is the probability that player $1$ places a label into some basket, which is $1-(1-p)^m$.
The probability of player $2$ being a winner is the probability that player $2$ places a label into some basket into which player $1$ doesn't place a label, which is $1-(1-p(1-p))^m$.
The probability of player $k$ being a winner is the probability that player $k$ places a label into some basket into which players $1$ through $k-1$ don't place a label, which is $1-\left(1-p(1-p)^{k-1}\right)^m$.
Thus the expected number of winners is
$ \sum_{k=1}^n\left(1-\left(1-p(1-p)^{k-1}\right)^m\right)\;. $
I don't see how to get a closed form without summation for this. If $n\gg m$, you can transform it into a sum up to $m$ so you have fewer to terms to sum:
$ \begin{align} \sum_{k=1}^n\left(1-\left(1-p(1-p)^{k-1}\right)^m\right) &= n-\sum_{k=1}^n\sum_{j=0}^m\binom mj(-p)^j(1-p)^{j(k-1)} \\ &= n-\sum_{j=0}^m\binom mj(-p)^j\frac{1-(1-p)^{jn}}{1-(1-p)^{j\hphantom n}}\;. \end{align} $