We have, $x^3-7 x^2+14 x-8=0$
Its roots are diminished by $k$.
$$
\begin{gathered}
\therefore(x-k)^3-7(x-k)^2+14(x-k)-8=0 \\
\Rightarrow x^3-3 x^2 k+3 x k^2-k^3-7\left(x^2-2 k x+k^2\right)+14 x \\
\Rightarrow x^3-x^2(3 k+7)+\left(3 k^2+14 k+14\right) \quad-8=0 \\
k^3-7 k^2-14 k-8=0
\end{gathered}
$$
This equation becomes.
$$
\begin{aligned}
y^3+p y-\frac{20}{27} & =0 \Rightarrow 3 k+7=0 \\
k & =\frac{-7}{3}
\end{aligned}
$$
Coefficient of $x$ is
$$
3 k^2+14 k+14=p
$$

$$
\begin{array}{ll}
\Rightarrow & \quad 3\left(-\frac{7}{3}\right)^2+14\left(\frac{-7}{3}\right)+14=p \\
\Rightarrow & p=\frac{49}{3}-\frac{98}{3}+14 \Rightarrow p=\frac{49-98+42}{3} \\
\Rightarrow & p=-\frac{7}{3}
\end{array}
$$