a
It is clear from the above multiplication, the degree of determinant of $B(x)$ can not be less than $4$ .
Let ${p_1}x = {a_1}{x^2} + {b_1}x + {c_1}$
${p_2}x = {a_2}{x^2} + {b_2}x + {c_2}$
and ${p_3}x = {a_3}{x^2} + {b_3}x + {c_3}$
where ${a_1},{a_2},{a_3},{b_1},{b_2},{b_3},{c_1},{c_2},{c_3}$ are real number.
$\therefore A\left( x \right) = \left[ {\begin{array}{*{20}{c}}
{{a_1}{x^2} + {b_1}x + {c_1}}&{2{a_1}x + {b_1}}&{2{a_1}}\\
{{a_2}{x^2} + {b_2}x + {c_2}}&{2{a_2}x + {b_2}}&{2{a_2}}\\
{{a_3}{x^2} + {b_3}x + {c_3}}&{2{a_3}x + {b_3}}&{2{a_3}}
\end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}}
{{a_1}{x^2} + {b_1}x + {c_1}}&{{a_2}{x^2} + {b_2}x + {c_2}}&{{a_3}{x^2} + {b_3}x + {c_3}}\\
{2{a_1}x + {b_1}}&{2{a_2}x + {b_2}}&{2{a_3}x + {b_3}}\\
{2{a_1}}&{2{a_2}x + {b_2}}&{2{a_3}}
\end{array}} \right]$
$ \times \left[ {\begin{array}{*{20}{c}}
{{a_1}{x^2} + {b_1}x + {c_1}}&{2{a_1}x + {b_1}}&{2{a_1}}\\
{{a_2}{x^2} + {b_2}x + {c_2}}&{2{a_2}x + {b_2}}&{2{a_2}}\\
{{a_3}{x^2} + {b_3}x + {c_3}}&{2{a_3}x + {b_3}}&{2{a_3}}
\end{array}} \right]$