Given a multiset of positive integers, its P-graph is the loopless graph whose vertex set consists of those integers, any two of which are joined by an edge if they have a common divisor greater than 1, that is, they are not relatively prime. There are 5,371,315,400 partitions of 130 of which 507,334 are partitions into six parts. I have been told that there is just one of these that can be uniquely recovered from its P-graph. Which is it?
I had originally found three P-graphs but, as OP correctly pointed out, two of them were invalid and coincided with multiple partitions. This was due to an error in my reasoning.
With the help of a computer, I found the solution can be represented diagramatically as follows:
Together with the corresponding partition, the diagram appears as follows
Explanation of method
1. List all partitions of length $6$.
2. Compute the adjacency matrix of the p-graph associated to each partition.
3. Split the set of unique adjacency matrices into those which appear more than once, $S$, and those that appear exactly once, $T$
4. Partition the set, $T$, into equivalence classes where two matrices are equivalent if they have the same element sum.
5. For each equivalence class $\Gamma$:
(i) For the first element, $m$, in the class, compute the list, $L$, of adjacency matrices related to that element by interchanging indices
(ii) Find the list, $l$, of all matrices in $\Gamma$ which appear in $L$.
(iii) If the length of $l$ is $1$ then $m$ is a candidate solution and we add it to a set $M$.
(iv) Remove all elements of $l$ from $\Gamma$.
(v) If $|\Gamma|> 0$, return to (i). If $|\Gamma|=0$, break.
6. For each element $m$ in $M$:
(i) Recompute the list of adjacency matrices $L_m$ related to $m$ by interchanging indices.
(ii) For each element of $L_m$, check if it is an element of $S$. If not for any element of $L_m$, then $m$ is a solution.