On the proof of NP-Hardness of the Cardinality Constrained Quadratic Knapsack Problem in Polyhedral Study of the Cardinality Constrained Knapsack Problem the authors prove that the Cardinality Constrained Knapsack Problem is NP-Hard by reducing PARTITION to it.

Besides, it's easy to see that the KP problem is a special case of the QKP.

How should one proceed to prove the NP-Hardness of the Cardinality Constrained Quadratic Knapsack Problem?

CCQKP:

$$max\ \sum_{i=1}^{n} \sum_{j=1}^{n} x_i x_j c_{ij}$$ 
$s.t$
$$\sum_{j=1}^{n} a_j x_j \leq C$$
$$\sum_{j=1}^{n}x_j = 1$$
$$0 \leq x_j \leq 1,\ j = 1,...,n$$
$$\sum_{j=1}^{n}z_j = k$$
$$z_j \in \{0,1\},\ j = 1,...,n$$

I'm aware that we usually talk about the hardness of Decision Problems even tough I'm formalizing the Optimization version of it.
The quadratic knapsack problem is NP-hard.  Therefore, the cardinality-constrained quadratic knapsack problem is NP-hard, too (as Yuval says, just don't use any cardinality constraints).

If problem X is a special case of problem Y, and problem X is NP-hard, then so is problem Y.  Now let X = CCQKP and Y = QKP.