Given,

$A,P,B$ are $3\times 3$ matrices.

And $\left|-B\right|=5\Rightarrow \left|B\right|=-5$

And $\left|B{A}^T\right|=15\Rightarrow \left|B\right|\left|A^T\right|=15\Rightarrow \left|A\right|=-3$

{as $\left|AB\right|=\left|A\right|\left|B\right| \& \left|A^T\right|=\left|A\right|$}

Now solving $\left|P^T\mathrm{AP}\right|=-27$ and same above property we get,

$\Rightarrow \left|P^T\right|\left|A\right|\left|P\right|=-27$

$\Rightarrow \left|P^T\right|\left|P\right|=9$

$\Rightarrow {\left|P\right|}^2=9$

$\Rightarrow \left|P\right|=\pm 3$

Then one of the values of $\left|P\right|$ is $3$