There exists such element. Indeed, if $g$ be a primitive root modulo $p$, exactly one of $g$ and $g+p$ is a primitive root module $p^2$ and the other isn't. Also, $g$ is a primitive root module $p$ such that isn't primitive root module $p^2$ iff ord$_pa=$ord$_{p^2}a=p-1$. therefore if $g$ be such element, $g^{-1}\pmod {p^2}$ is also such element. furthermore if $p\geq3$ and $g\equiv g^{-1}\pmod{p^2}$ then $g^2\equiv 1\pmod {p^2}$ and thus $g$ can not be a primitive root module $p$. so the answer is $1$.

This was exactly my solution on exam, but remember, you should check the case $p=2$.

Another Solution: Define $f(n) = \prod_{ord_{p^2}(d)=p-1}(-d)$ and $F(n) = \prod_{d|n}f(d)$ now compute F and use $f(n) = \prod_{d|n}F(d)^{\mu{(\frac{n}{d})}}$ (use the fact the problem asks for the product of the numbers whose order is p-1 modulo p^2)