Will the addition happen across both double bonds? Ordinarily you add H and OH syn and the OH anti-Markovnikov to one double bond
Hypothetical product:
$\ce{BH_3} $ can attack any of the two double bonds. Now it all depends on how many moles are being reacted in the vessel. Let us suppose there is only one mole. In the below process there is formation of chiral carbon and hence there is a pair of enantiomer.
Now we consider the other one.
Here there is no chiral center and hence no optical isomerism. Hope this was helpful to you....
I personally feel that the addition should be on the left double bond as it has least number of neighbouring alpha hydrogens to give it a sort of hyperconjugative delocalisation ,and thus is relatively more reactive than the double bond at the right extreme.