(Moved from comments) Here's an idea for getting a constant factor approximation, assuming P and Q satisfy the triangle inequality. I thought it might give a 2-approximation, but all I can prove right now is an approximation ratio of 4.
(1) In the problem as stated, the weight of edge $pq$ in the combined graph (after the correspondence $p$–$p'$ and $q$–$q'$ is determined) is $\max\{P(pq),Q(p'q')\}$. Instead, let's use $P(pq)+Q(p'q')$. This loses at most a factor of two but makes the problem easier to describe: we are now trying to find a spanning tree in $P$, and an isomorphic spanning tree in $Q$, with minimum total weight. The correspondence between $P$ and $Q$ is then given by the isomorphism between these two trees.
(2) In $P$, find a minimum spanning tree, and use the path-doubling Euler tour technique to find a path with at most twice the weight. Do the same thing independently in $Q$. The result is two isomorphic trees (both paths) that are separately at most twice the weight of their graph's MSTs, and therefore at most twice the cost of the solution to the minimum isomorphic spanning tree problem, and four times the weight of the original problem.
(3) The original problem is NP-complete, by a reduction from Hamiltonian path. Let $P$ be defined from a graph in which you wish to test the existence of a Hamiltonian path; define $P(pq)=1$ when $pq$ is an edge in $P$ and $2$ when $pq$ is not an edge. Let $Q$ be defined in exactly the same way from a path graph. Then there is a solution of total cost $n-1$ if and only if the graph from which $P$ was defined has a Hamiltonian path. Probably this can also be used to prove inapproximability below some fixed constant.
