the two half-reactions, balanced as if in acidic solution:
| 12e¯ + 12H+ + P4 | ---> | 4PH3 |
| 12H2O + P4 | ---> | 4HPO32¯ + 20H+ + 12e¯ |
convert to basic solution:
| 12e¯ + 12H2O + P4 | ---> | 4PH3 + 12OH¯ |
| P4 + 20OH¯ | ---> | 4HPO32¯ + 8H2O + 12e¯ |
In the second half-reaction, after adding 20 OH¯, you had 12H2O on the left side and 20H2O on the right. So 12H2O were eliminated, leaving the second half-reaction as displayed.
Note that the electrons are already equal and add the two half-reactions. Remember to eliminate eight waters and 12 hydroxides from each side to get the final answer:
| 4H2O + 2P4 + 8OH¯ | ---> | 4PH3 + 4HPO32¯ |