Show that the distribution of $X$ is like this:

Assume $0<\sigma\leq k$. First, we show $\text{supp}\{X\}\subseteq\{\mu,\mu+k,\mu-k\}.$

Lemma: Suppose $Z\geq 0,a>0$ and $P(Z\geq a)=E[Z]/a.$ Then $\text{supp}\{Z\}\subseteq\{0,a\}$

Proof: Note ${\bf 1}_{Z\geq a}\leq Z/a$ for $Z\geq 0$ with equality only holding at $Z=0,a.$ Hence $P(Z\geq a)=E[{\bf 1}_{Z\geq a}]= E[Z/a]$ only if $Z=0$ or $Z=a$ a.s.$\square$

Now we are given $\frac{1}{k^2/\sigma^2}=P(|X-\mu|\geq k)=P\left(\frac{(X-\mu)^2}{\sigma^2}\geq \frac{k^2}{\sigma^2}\right),$ so the claimed three-point support follows immediately from our lemma.

Now we simply need to find the mass on each point. Note $$P(X=\mu)=P(|X-\mu|< k)=1-P(|X-\mu|\geq k)=1-\frac{\sigma^2}{k^2}.$$

Noting the three masses sum to unity, and $\mu+k$ and $\mu-k$ are equidistant from mean $\mu$, the other masses follow:

$$P(X=\mu+k)=P(X=\mu-k)=\frac{\sigma^2}{2k^2}.$$