There are four stereocentres in hexoaldoses giving rise to a set of $2^4 = 16$ different molecules with the same connectivity; you can group these to eight pairs of enantiomers (perfect mirror images of each other). The enantiomers are labelled D and L depending on the orientation of the stereocentre which is farthest from the aldehyde group. In linear orientation, these are the following molecules:
(All images taken from Wikipedia; all D-isomers.)
In the order of reading these are allose, altrose, glucose, mannose, gulose, idose, galactose and talose. If you compare glucose (3rd in the first row) with galactose (3rd in the second row) you will realise that they differ only in the orientation of the hydroxy group in 4-position. The posh terminology is epimerism — D-glucose is the C4 epimer of D-galactose and vice-versa.
You asked about two labels where the sugar is preceded either by 4-dehydro-6-deoxy or 4-keto-6-deoxy. The agree in the 6-deoxy part which means that the oxygen on C6 is missing; you will find a methyl group $\ce{CH3}$ there instead. The labels 4-keto and 4-dehydro may also be confusing. The first says that a keto group is there instead of the hydroxy group, while the second says that two hydrogens have been removed from that position. The result is the same, instead of having $\ce{CHOH}$ as fourth carbon we have $\ce{C=O}$. This also deprives us of the stereocentre (a carbonyl group is never stereogenic).
Putting all of this together: Yes, the two are identical. If you start from the C4-epimers and remove the stereocentre at C4 it does not matter which you started from. Both galactose and glucose are common sugars so both choices are understandable (even though glucose is probably known even a tad more). I would, however, strongly object agains 4-keto-D-talose in the place of 4-keto-D-mannose (mannose and talose being C4-epimers) since mannose is much more known than talose.
