= Solution
About a fixed origin, the total <momentum> and <angular momentum> are
$$
\mathbf P=\sum_i m_i\mathbf v_i,\qquad \mathbf L=\sum_i\mathbf x_i\times m_i\mathbf v_i.
$$
A time-independent translation $\mathbf x'_i=\mathbf x_i-\mathbf b$ leaves every <velocity> unchanged. Therefore
$$
\boxed{\mathbf P'=\mathbf P,\qquad \mathbf L'=\mathbf L-\mathbf b\times\mathbf P}.
$$
The <cross product> $\mathbf b\times\mathbf P$ is perpendicular to $\mathbf P$, giving the invariant
$$
\boxed{\mathbf L'\cdot\mathbf P'=\mathbf L\cdot\mathbf P}.
$$
The other requested transformation follows from the vector triple-product identity:
$$
\boxed{\mathbf L'\times\mathbf P'=\mathbf L\times\mathbf P+|\mathbf P|^2\mathbf b-(\mathbf b\cdot\mathbf P)\mathbf P}.
$$
The actual PDF asks about $\mathbf L\times\mathbf P$; the TeX's occurrence of $\mathbf F$ is a transcription error.
If $\mathbf P=0$, the translation term vanishes and <angular momentum> is independent of origin. If $\mathbf P\ne0$, choose
$$
\boxed{\mathbf b=\frac{\mathbf P\times\mathbf L}{|\mathbf P|^2}+\lambda\mathbf P,\qquad \lambda\in\mathbb R}.
$$
Then $\mathbf b\times\mathbf P=\mathbf L-(\mathbf L\cdot\mathbf P)\mathbf P/|\mathbf P|^2$, and so
$$
\mathbf L'=\frac{\mathbf L\cdot\mathbf P}{|\mathbf P|^2}\mathbf P.
$$
Thus the new <angular momentum> is parallel to the total <momentum>. The freedom along $\mathbf P$ gives the <central axis of a momentum system>. If the invariant <dot product> is zero, the new <angular momentum> is the zero vector. All expressions assume a translated fixed origin, not a moving reference frame.
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