Option A

Note that in the field of characteristic $2$, \[ f(x) = x^4 + x \implies f'(x) = 4x^3 + 1 = 1, \] which can never be zero. Thus, in the field of characteristic $2$, the given polynomial does not have any root with multiplicity greater than $1$.

Option B

Again in the field of characteristic $3$, the polynomial will be \[ f(x) = x^4 + x \implies f'(x) = 4x^3 + 1. \] It is clear that \[ f(2) = 0 \quad \text{ and } \quad f'(2) = 0 \] in $\mathbb{F} _3[x]$. Thus, in the field of characteristic $3$, the given polynomial has a root of multiplicity greater than $1$.

Option C

Now consider the field with characteristic $5$. Then we have \[ f(x) = x^4 + x + 1 \implies f'(x) = 4x^3 + 1. \] Note that \[ f(1) \neq 0, f(2)\neq 0 \text{ and } f(3) = 0 \text{ but } f'(3) \neq 0. \] Thus, for $p = 5$ also we don't have any root with multiplicity bigger than $1$.

Option D

Similarly, for $p = 7$, note that $f(x) = x^4 + x + 6 $ does not have a root in $\mathbb{Z} _7$ and thus for $p = 7$, we don't have any root with multiplicity bigger than $1$.