I underestimated one fact that I have still have to check |p|=|q|=|r|=|s|, if p is not equal to q then you check whether p is equal to s, and npda is possible for this considering first satisfies their respective intermediate constrains, but then later I realized it is accepting strings where |p|=|q|=|r|=|s|. So I tried to go in details by taking one language at a time.
$p^{n}q^{m}r^{m}s^{n}$ , and n!= m but I didn't found any way to accept this as I first need to check whether n==m or not, which may or may not empty my stack. If I do so I won't be having considerable amount of p's to check with s. and hence it won't be CFL.
Thank you Shaik for pointing out a mistake.