Лиези:
1. Строка в столбец
select regexp_split_to_table(permit_type,',') as code from food_perfect_factor a where a.permit_number = 'SC20152272518392'
2. Столбец для строки
select string_agg(a.code,',') from (select regexp_split_to_table(permit_type,',') as code from food_perfect_factor a where a.permit_number = 'SC20152272518392') a