= Solution
For $f_z(y)=f(yz)$, the substitution $u=yz$ gives
$$
\|f_z\|_p^p=z^{-1}\|f\|_p^p,
\qquad
\|f_z\|_p=z^{-1/p}\|f\|_p.
$$
Homogeneity of $K$ and the substitution $x=yz$ give
$$
Tf(y)=\int_0^\infty K(yz,y)f(yz)y\,dz
=\int_0^\infty K(z,1)f_z(y)\,dz.
$$
Apply part (a), or equivalently the <Minkowski integral inequality>, to this <integral> and use the preceding scaling identity:
$$
\begin{aligned}
\|Tf\|_p
&\leq\int_0^\infty |K(z,1)|\,\|f_z\|_p\,dz\\
&=\|f\|_p\int_0^\infty |K(z,1)|z^{-1/p}\,dz\\
&=\|f\|_p.
\end{aligned}
$$
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