Aqui está uma opção. Leia comentários dentro do código. Dados de amostra nas linhas #1 - 13; consulta começa na linha #14.
SQL> with
2 expl (id, name) as
3 (select 1, 'car' from dual union all
4 select 2, 'bus' from dual union all
5 select 3, 'BB' from dual union all
6 select 4, 'SB' from dual union all
7 select 5, 'Ba' from dual union all
8 select 6, 'PA' from dual union all
9 select 7, 'HB' from dual union all
10 select 8, 'G' from dual
11 ),
12 temp (col) as
13 (select '1,4,7,8' from dual),
14 -- split COL to rows
15 spl as
16 (select regexp_substr(col, '[^,]+', 1, level) val,
17 level lvl
18 from temp
19 connect by level <= regexp_count(col, ',') + 1
20 )
21 -- join SPL with EXPL; aggregate the result
22 select listagg(e.name, ',') within group (order by s.lvl) result
23 from expl e join spl s on s.val = e.id;
RESULT
--------------------------------------------------------------------------------
car,SB,HB,G
SQL>
