= Solution
The free part of the abelianization is generated by $[y]$, so the fibration class sends $y\mapsto1$ and $x\mapsto0$. Applying <Fox calculus> to the relator and then substituting $x=1$, $y=t$ gives
$$
\left.\frac{\partial r}{\partial x}\right|_{x=1,y=t}
=4-t-t^{-1}.
$$
This has breadth two. Since $K$ is assumed fibered, the breadth is twice the genus, so its fiber is a genus-one surface with one boundary component:
$$
\boxed{\Sigma\cong T^2\setminus\operatorname{int}D^2.}
$$
The monodromy can also be read directly from the relator. Put $x_i=y^ixy^{-i}$. Reidemeister rewriting gives
$$
x_0^2x_{-1}^{-1}x_0x_1^{-1}x_0=1,
\qquad
x_1=x_0^3x_{-1}^{-1}x_0.
$$
Thus the kernel is freely generated by $(x_{-1},x_0)$, and conjugation by $y$ sends
$$
x_{-1}\mapsto x_0,
\qquad
x_0\mapsto x_0^3x_{-1}^{-1}x_0.
$$
On first homology, in the ordered basis $([x_{-1}],[x_0])$, a representative is
$$
\boxed{\phi_*=
\begin{pmatrix}
0&-1\\
1&4
\end{pmatrix}.}
$$
Its characteristic polynomial is $t^2-4t+1$, agreeing up to the unit $-t$ with $4-t-t^{-1}$.
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