Consensus clustering using set union
The problem is NP-hard. Proof is by reduction from the following problem:

Given a tripartite graph $G$ with $N$ vertices in each part, are there $N$ vertex-disjoint triangles in $G$?

Here's the reduction: given an instance $G$ of the above problem, let $A_1$, $A_2$, $A_3$ denote the set of vertices in each part of $G$, and $E_{ij}$ be the set of edges between $A_i$ and $A_j$. Also, number the vertices in each part by $1,\dotsc,N$.

We construct an instance of your problem with $n = |E(G)| + M$, where $M$ is a large number (say, $M = 10 |E(G)|$), and $p = N + 1$.  The first $|E(G)|$ elements of $\{1,\dotsc,n\}$ correspond to the edges of $G$.  The partition $P$ is defined as follows: $P_i$ for $i=1,\dotsc,N$ is the set of edges that have the $i$th vertex of $A_1$ as one of their endpoints. Clearly these sets are disjoint and their union is $E_{1,2} \cup E_{1,3}$. $P_{N+1}$ is everything else, i.e., $E_{2,3} \cup \{|E(G)| + 1, \dotsc, |E(G)| + M\}$. Similarly, we define $Q$ using $A_2$ instead of $A_1$, and $R$ using $A_3$ instead of $A_1$. 

Now, we claim is that this instance has a solution of cost at most $3|E(G)| - 3N + M$ if and only if $G$ has $N$ disjoint triangles. To see this, first notice that since $M$ is large, any solution that has cost less than $2 M$ must map $P_{N+1}$ to $Q_{N+1}$ and $R_{N+1}$. This already accounts for $|E(G)| + M$ of the total cost, so we are left with $2 |E(G)| - 3 N$. Now, note that the intersection of each $P_i$ and each $Q_j$ is at most one (and similarly for $P_i$ and $R_k$, and also for $Q_j$ and $R_k$). So, the objective function is minimized if all these intersections can be 1 at the same time. This corresponds to $N$ disjoint triangles in $G$.