If there would be a polynomial time algorithm for your problem, it could be used to solve the NP-hard recognition problem in polynomial time by just giving the input to it and checking if its output is correct. Therefore the problem you pose is NP-hard in the sense that if it admits polynomial time algorithm, then P=NP.
(Here we assume that the input is some well-known encoding of a graph, and therefore the restriction that the input must be an unit disk graph doesn't really make the problem at all easier. Maybe there could be some other encodings in which only unit disk graphs could be represented, but formulating them would be another topic.)
Edit: I'll try to formalize the question and the answer more:
Formalization of the question:
Is there a Turing machine $M$ and a polynomial $p(n)$, such that if the input of $M$ is a unit disk graph $G$ with $n$ vertices encoded as an adjacency matrix, then $M$ terminates in at most $p(n)$ steps, and outputs an unit disk configuration of $G$? In particular, the machine $M$ is allowed to have undefined behavior if the input is not an unit disk graph encoded as adjacency matrix.
Answer: Suppose that there is such Turing machine $M$ and polynomial $p(n)$. Now, we can solve unit disk recognition problem in polynomial time as follows: Given a graph $G$ with $n$ vertices, run $M$ with $G$ as input for $p(n)$ steps. If $M$ does not terminate in $p(n)$ steps, output NO. If $M$ terminates, test if the output of $M$ is an unit disk configuration of $G$. If it is, output YES and otherwise NO. Now we have solved an NP-hard problem in polynomial time, so we have a contradiction if we assume $P \neq NP$.