The probability that $\ge j$ problems are solved is the probability that:
- At least one student solves one of the problems
- At least one of the remaining students solves a second problem
$\qquad \qquad \vdots$
- At least one of the remaining students solves a $j^{\text{th}}$ problem
And these events are all independent. And the probability that any given student chooses and solves any given problem is $\dfrac{p}{k}$, so we obtain:
$\mathbb{P}(X \ge j) = dk\dfrac{p}{k} \cdot (d-1)(k-1)\dfrac{p}{k} \cdot (d-j+1)(k-j+1) \dfrac{p}{k}$
which is precisely $\displaystyle \dfrac{p^j}{k^j} \binom{d}{j} \binom{k}{j}$.
Hence we have
$\displaystyle \mathbb{E}(X) = \sum_{j=0}^k \mathbb{P}(X \ge k) = \sum_{j=0}^k \dfrac{p^j}{k^j} \binom{d}{j} \binom{k}{j}$
I'm not sure how you could simplify this, but no doubt there is a way.
Another way of visualising it is this: you have a $d \times k$ grid, where the $d$ rows represent students and the $k$ columns represent the problems. You select one square from each row, and you need to work out the expected number of columns which have at least one square chosen from them.