= Solution
For every nonzero $v\in\mathbb R^{n-p}$, full column rank of $L$ gives $Lv\ne0$, and hence
$$
v^TL^TLv=\|Lv\|^2>0.
$$
Thus the <Gram matrix> $L^TL$ is a <positive-definite matrix> and is invertible. Using $BA=0$ and $A=LL^T$,
$$
0=BLL^TL(L^TL)^{-1}=BL,
$$
so
$$
\boxed{L^TL\text{ is positive definite},\qquad BL=0.}
$$
A <linear image of a multivariate normal vector> is again multivariate normal. Therefore the stacked vector has mean $(B\mu,L^T\mu)^T$ and <covariance>
$$
\sigma^2\begin{pmatrix}BB^T&BL\\L^TB^T&L^TL\end{pmatrix}=\sigma^2\begin{pmatrix}BB^T&0\\0&L^TL\end{pmatrix}.
$$
In particular,
$$
\boxed{Z\sim N_n\!\left(\begin{pmatrix}B\mu\\L^T\mu\end{pmatrix},\ \sigma^2\begin{pmatrix}BB^T&0\\0&L^TL\end{pmatrix}\right).}
$$
The two diagonal blocks are <positive-definite> by the rank assumptions. The two blocks are jointly Gaussian and have zero cross <covariance>. The fact that <uncorrelated jointly Gaussian variables are independent> makes $BY$ and $L^TY$ independent. Since
$$
Y^TAY=Y^TLL^TY=\|L^TY\|^2
$$
is a measurable function of the second block, independence is preserved under this transformation. This proves <Gaussian independence of linear and quadratic statistics>:
$$
\boxed{BY\text{ and }Y^TAY\text{ are independent}.}
$$
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