NP-hardness of an optimization problem
Well, here's a possible solution:

The reduction will be from 3SAT.

Input: $m$ DNF clauses $(\varphi_1,\ldots,\varphi_m)$ over $n$ variables $(x_1,\ldots,x_n)$.

Reduction:
Create a set of items composed of two items for every variable: $x_i, \overline x_i$, corresponding to a $True$ assignment to either the variable $x_i$ or its negation, plus one auxiliary item $t$.
Let the price of all items $\{x_i,\overline x_i\}_{i=1,…,n}$ be $1$, and the price of $t$ be $1.5$.

Create two sets of consumers:

Set 1: Validity rankings: this set of consumers will encode the validity constraints on the assignments to $x_1,\ldots,x_n$. Namely, that exactly one out of every $\{x_i,\overline x_i\}$ is set to $True$ (i.e., taken by the algorithm).
For every $i=1,\ldots,n$ create four partial rankings:

$\sigma_{i1}: x_i\succ t\\
\sigma_{i2}: \overline x_i \succ t \\
\sigma_{i3}: x_i \succ \overline x_i \\
\sigma_{i4}: x_i \succ \overline x_i$

Taking $t$ can never hurt, so we assume that it's always chosen. If both $x_i$ and $\overline x_i$ are selected, we get a payoff $4$. If none of them is selected, we get a payoff of $3$. If one of them is chosen, we get a payoff of $4.5$.

Set 2: Clause rankings.
For each clause of the form $\varphi_j=\ell_{j1} \vee \ell_{i2} \vee \ell_{j3}$, we create a ranking: $\sigma_j: \ell_{j1} &gt; \ell_{j2} &gt; \ell_{j3}$. If $\varphi_j$ is satisfied, it means that at least one of the items corresponding to $\ell_{j1}, \ell_{j2}$, and $\ell_{j3}$ is selected, which gives an extra payoff of $1$ from ranking $\sigma_j$.

Now, the 3CNF formula is satisfiable if and only if there is a set of items that gives a payoff of $m+4.5n$.