Let's first choose the three columns. There are four cases,
- all columns even
- two columns even, one odd
- one column even, two odd
- all columns odd
It either of cases (1) and (4), we have four rows we could choose for the lowest numbered column, three for the next lowest numbered column, and two for the highest numbered column. This gives $ 2\cdot\binom{4}{3}\cdot(4\cdot3\cdot2) $ selections.
Now look at case (2). We have four row choices for the lower even column, and three for the higher even column. We have four row choices for the odd column. Case (3) is similar, so we get $ 2\binom{4}{2}\binom{4}{1}\cdot(4\cdot3\cdot4) $ selections.