How to insert duplicate records with different id if its not present with that id

If you are saying that every value of [name] must have an entry for all possible values of [pdcode] then...

1. Determine what all the possible combinations of [name] and [pdcode] are required. If you assume that the table contains at least one record for each [name] AND at least one record for each [pdcode] you can do this by selecting the DISTINCT[^] values of each and using CROSS APPLY[^]

2. You can put those results into a Common Table Expression (CTE) - see Common Table Expressions(CTE) in SQL SERVER 2008[^]

3. If you use the results from the CTE as the left table (i.e. what we want) and LEFT OUTER JOIN[^] to what we already have on the table you can filter the results for where pdcode from the original table is NULL - i.e. the list of "missing" rows

4. You can then just INSERT INTO[^] those values into the original table with a default value for the amount

It's a lot simpler than it sounds...
- Create Test data:

create table joemens
(
    [name] int,
    pdcode int,
    amount int
)

insert into joemens values
(1,20,5),
(1,30,10),
(2, 20, 199),
(3,30,40)

- Identify all the values we need

SELECT * FROM
(select distinct pdcode from joemens) A
CROSS APPLY (select distinct [name] from joemens) B

gives results of

pdcode name
20     1
20     2
20     3
30     1
30     2
30     3

- Stick it into a CTE and identify the gaps

;WITH CTE AS(
    SELECT * FROM
    (select distinct pdcode from joemens) A
    CROSS APPLY (select distinct [name] from joemens) B
)
select * -- CTE.[Name], CTE.pdcode, 0
from CTE
left outer join joemens J on J.[name]=CTE.[name] AND J.pdcode=CTE.pdcode

which results in

20	1	1	20	5
20	2	2	20	199
20	3	NULL	NULL	NULL
30	1	1	30	10
30	2	NULL	NULL	NULL
30	3	3	30	40

- pull it all together - we want to insert the rows where the NULLs appear

;WITH CTE AS(
    SELECT * FROM
    (select distinct pdcode from joemens) A
    CROSS APPLY (select distinct [name] from joemens) B
)
INSERT INTO joemens
select CTE.[Name], CTE.pdcode, 0
from CTE
left outer join joemens J on J.[name]=CTE.[name] AND J.pdcode=CTE.pdcode
WHERE J.pdcode IS NULL