Irska

When $p=2$, $n=63$. When $p=3$, $n=103$. When $p=5$, $n=567$.

$$\begin{array}{rl}(p^2+2{)}^2-9(p^2-7)& =p^4-5p^2+4+63\\ & =(p^2-1)(p^2-4)+63\\ & =(p-1)(p+1)(p-2)(p+2)+63\\ & =(p-2)(p-1)(p+1)(p+2)+63\end{array}$$

When $p\ne 3$, $(p-2)(p-1)$ and $(p+1)(p+2)$ are both multiples of 3. Thus $n=(p^2+2{)}^2-9(p^2-7)$ is a multiple of 9 and so the digit sum is a multiple of 9. When $n=3$, the digit sum is 4. Thus this is the smallest value that can be obtained.