Given: ${\mathit{x}}^2+\left(\mathit{p}-3\right)\mathit{x}+\mathit{p}=0$
$\because$ Roots are real and equal, then
${\mathit{b}}^2-4\mathit{a}\mathit{c}=0$
Here we compare the coefficients of $\mathit{a},\mathit{b}$ and $\mathit{c}$ with the equation $\mathit{a}{\mathit{x}}^2+\mathit{b}\mathit{x}+\mathit{c}=0$ .

Now putting the values of $\mathit{a},\mathit{b}$ and $\mathit{c}$ in given equation
${\left(\mathit{p}-3\right)}^2-4\times 1\times \mathit{p}=0$
${\mathit{p}}^2+9-6\mathit{p}-4\mathit{p}=0$
${\mathit{p}}^2+9-10\mathit{p}=0$
${\mathit{p}}^2-10\mathit{p}+9=0$
${\mathit{p}}^2-9\mathit{p}-\mathit{p}+9=0$
$\mathit{p}\left(\mathit{p}-9\right)-1\left(\mathit{p}-9\right)=0$
$\Rightarrow \left(\mathit{p}-9\right)\left(\mathit{p}-1\right)=0$
Hence, $\mathit{p}=9$ or 1