If $\left(x-1\right)$  is a factor of $x^4-\left(p-5\right)x^3-\left(4p-6\right)x^2+\left(3p-9\right)x+6,$  then the value of $p$ is

The given expression is $x^4-\left(p-5\right)x^3-\left(4p-6\right)x^2+\left(3p-9\right)x+6,$

And  $x-1$   is a factor$\Rightarrow 1$  is the zero of the polynomial

So substitute $x=1$ in given expression $\Rightarrow {\left(1\right)}^4-\left(p-5\right){\left(1\right)}^3-\left(4p-6\right){\left(1\right)}^2+\left(3p-9\right)\left(1\right)+6=0$

$\Rightarrow 1-p+5-4p+6+3p-9+6=0$

$\Rightarrow p=\frac{9}{2}$