$\begin{aligned} & =\left[\begin{array}{rr}64+10 & -40+20 \\ -16+8 & 10+16\end{array}\right] \\ & =\left[\begin{array}{ll}74 & -20 \\ -8 & 26\end{array}\right] \\ & 4 A=\left[\begin{array}{rr}-32 & 20 \\ 8 & 16\end{array}\right] \\ & -p l=\left[\begin{array}{cc}-p & 0 \\ 0 & -p\end{array}\right] \\ & \end{aligned}$
Since, the matrix $A$ satisfies the equation
$$
\begin{aligned}
& x^2+4 x-p=0, \text { then } \\
& A^2+4 A-p l=0 \\
& \Rightarrow {\left[\begin{array}{rr}
74 & -20 \\
-8 & 26
\end{array}\right]+\left[\begin{array}{rr}
-32 & 20 \\
8 & 16
\end{array}\right]+\left[\begin{array}{cc}
-p & 0 \\
0 & -p
\end{array}\right] } \\
& \Rightarrow {\left[\begin{array}{ll}
74-32-p & -20+20+0 \\
{\left[\begin{array}{cc}
0 & 0 \\
-8+8+0 & 26+16-p
\end{array}\right]=\left[\begin{array}{ll}
0 & 0
\end{array}\right]} \\
0 & 0
\end{array}\right] } \\
& \Rightarrow {\left[\begin{array}{cc}
42-p & 0 \\
0 & 42-p
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right] }
\end{aligned}
$$
On comparing, we get
$$
42-p=0 \Rightarrow p=42
$$