= Solution
Suppose first that $x$ is a <purely inseparable algebraic element> over $K$, so
$$
x^{p^N}=y\in K
$$
for some $N\geq0$. Then $m_x$ divides $T^{p^N}-y$. Over an algebraic closure the <Frobenius endomorphism> is injective, so this polynomial has only one distinct root:
$$
T^{p^N}-y=(T-x)^{p^N}.
$$
Repeatedly use the zero-derivative criterion to write
$$
m_x(T)=q(T^{p^n})
$$
with $q'\ne0$. The polynomial $q$ is irreducible and therefore separable by part (a), but all its roots are powers of the single root $x$. Hence $q$ has degree one. Since $m_x$ is monic,
$$
m_x(T)=T^{p^n}-a
$$
for some $a\in K$.
Conversely, if the minimal polynomial has this form, then
$$
x^{p^n}=a\in K,
$$
so $x$ is purely inseparable. Thus the <minimal polynomial of a purely inseparable element> is exactly
$$
\boxed{m_x(T)=T^{p^n}-a,\qquad n\geq0,\ a\in K.}
$$
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