= Solution
Use the braid $B=(\sigma_1\sigma_2^{-1})^2$ from part (c). Put meridians $a,b,c$ around its three strands at one end. At a positive crossing, the <Wirtinger presentation> relations induce the <Artin automorphism of a braid>
$$
f:\ a\mapsto aba^{-1},\quad b\mapsto a,\quad c\mapsto c.
$$
For the inverse crossing on strands two and three, they give
$$
g:\ a\mapsto a,\quad b\mapsto c,\quad c\mapsto c^{-1}bc.
$$
The first formula is the usual crossing conjugation, and the second is its inverse on the relevant pair. Following one block gives $h=f\circ g$:
$$
h(a)=aba^{-1},\qquad h(b)=c,\qquad h(c)=c^{-1}ac.
$$
The four-crossing braid induces $h^2$. Closing it identifies each ending meridian with its starting meridian. Applying the <Seifert-van Kampen theorem> to the crossing regions and the closing arcs therefore gives
$$
\pi_1(S^3\setminus K)=\langle a,b,c\mid a=h^2(a),\ b=h^2(b),\ c=h^2(c)\rangle,
$$
where
$$
\begin{aligned}
h^2(a)&=aba^{-1}ca b^{-1}a^{-1},\\
h^2(b)&=c^{-1}ac,\\
h^2(c)&=c^{-1}a^{-1}c\,aba^{-1}c^{-1}ac.
\end{aligned}
$$
Both elementary crossing automorphisms preserve $abc$. Hence $h^2(abc)=abc$, and the first two closure relations imply the third: after substitution $abc=ab\,h^2(c)$, so cancellation gives $c=h^2(c)$.
The first remaining relation can be solved without adding any relation:
$$
a=aba^{-1}ca b^{-1}a^{-1}quad\Longleftrightarrow\quad c=ab^{-1}aba^{-1}.
$$
Substitute this expression into $b=c^{-1}ac$. Free cancellation and multiplying the two sides by inverse words give
$$
ab^{-1}a^{-1}ba=bab^{-1}a^{-1}b.
$$
These are <Tietze transformations>, eliminating a generator defined by one relation and retaining the other. Thus the <figure-eight knot group> has the requested one-relator presentation
$$
\boxed{\pi_1(S^3\setminus K)\cong
\langle a,b\mid ab^{-1}a^{-1}ba=bab^{-1}a^{-1}b\rangle.}
$$
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