Is there a necessary and sufficient condition for a linearly ordered topological space to be pseudo-metrizable? (by a pseudo-metric, I mean a map $X\times X\rightarrow\mathbb{R}$ in which all the metric axioms are satisfied except that the distance between two distinct points may be zero) 


A negative answer is, of course, still acceptable; I was wondering if there is an analogue of Urysohn's metrization theorem with respect to pseudo-metrics in linearly ordered spaces.



The pseudo-metric and linear orders
As Joel Hamkins observed above, a pseudometric may generate the topology of a linearly ordered topological space only if it is a metric. 


A necessary and sufficient condition for a linearly ordered topological space $X$ to be metrizable is that is has a $G\_\delta$-diagonal (that is, the set $\Delta=\{(x,x):x\in X\}$ is a $G\_\delta$ subset of $X\times X$ ). See the article by Lutzer in Encyclopedia of General Topology for further references, this article is available at the web-page of the author at: 
<http://www.resnet.wm.edu/~djlutz/Encyclopedia050602.pdf> . 
The same article also gives a condition due to Faber for the metrizability of generalized ordered spaces (equivalently subspaces of linearly ordered topological spaces): Existence of a dense, $\sigma$-closed-discrete subset satisfying an additional condition listed there. 


