Conditional Count on a field

IIF is not a standard SQL construct, but if it's supported by your database, you can achieve a more elegant statement producing the same result:

SELECT JobId, JobName,

COUNT(IIF (Priority=1, 1, NULL)) AS Priority1,
COUNT(IIF (Priority=2, 1, NULL)) AS Priority2,
COUNT(IIF (Priority=3, 1, NULL)) AS Priority3,
COUNT(IIF (Priority=4, 1, NULL)) AS Priority4,
COUNT(IIF (Priority=5, 1, NULL)) AS Priority5

FROM TableName
GROUP BY JobId, JobName

Using COUNT instead of SUM removes the requirement for an ELSE statement:

SELECT jobId, jobName,
    COUNT(CASE WHEN Priority=1 THEN 1 END) AS Priority1,
    COUNT(CASE WHEN Priority=2 THEN 1 END) AS Priority2,
    COUNT(CASE WHEN Priority=3 THEN 1 END) AS Priority3,
    COUNT(CASE WHEN Priority=4 THEN 1 END) AS Priority4,
    COUNT(CASE WHEN Priority=5 THEN 1 END) AS Priority5
FROM TableName
GROUP BY jobId, jobName

I think you may be after

select 
    jobID, JobName,
    sum(case when Priority = 1 then 1 else 0 end) as priority1,
    sum(case when Priority = 2 then 1 else 0 end) as priority2,
    sum(case when Priority = 3 then 1 else 0 end) as priority3,
    sum(case when Priority = 4 then 1 else 0 end) as priority4,
    sum(case when Priority = 5 then 1 else 0 end) as priority5
from
    Jobs
group by 
    jobID, JobName

However I am uncertain if you need to the jobID and JobName in your results if so remove them and remove the group by,