Solution:

Note that $A$ is the orthocenter of triangle $P_{2} P_{3} P_{4}$ by projecting the givens about $H$.

Next, let $P_{2} A$ intersect $P_{3} P_{4}$ at $C$ and $\left(P_{2} P_{3} P_{4}\right)$ at $D$. Then $A D=2 A C$.

Note that $H$ is the orthocenter of triangle $P_{1} P_{2} C$, with $A$ being the foot of the altitude from $P_{1}$. Thus $A P_{2} \cdot A C=A H \cdot A P_{1}$.

Finally, by Power of a Point at $A$,
$$A P_{1} \cdot A B=A P_{2} \cdot A D=2 A P_{2} \cdot A C=2 A H \cdot A P_{1}$$
Thus $A B=2 A H$ which gives $H B=3 H A$, as desired.

Solution:

We use vectors with $H$ as the origin. Then $P_{i} \cdot\left(P_{j}-P_{k}\right)=0$ from the definition of $H$. Thus $P_{i} \cdot P_{j}=P_{i} \cdot P_{k}$ for any distinct $i, j, k$. In particular, the quantity $P_{i} \cdot P_{j}$ is a constant over all distinct choices of $i, j$. Let this value be $D$. Let
$$O' = \frac{1}{2}\left(P_{1}+P_{2}+P_{3}+P_{4}\right)$$
We claim $O' = O$. To see this, note that
$$\left(O' - P_{1}\right) \cdot\left(O' - P_{1}\right)=\frac{1}{4}\left(P_{2}+P_{3}+P_{4}-P_{1}\right) \cdot\left(P_{2}+P_{3}+P_{4}-P_{1}\right)$$
will be symmetric after expanding. Thus $O'$ is equidistant from the $P_{i}$'s, which means $O' = O$.

Now, the projection of $O$ onto $P_{1} H$ is $\frac{O \cdot P_{1}}{P_{1} \cdot P_{1}} P_{1}$, so
$$B = P_{1}\left(\frac{\left(2 O - P_{1}\right) \cdot P_{1}}{P_{1} \cdot P_{1}}\right) = P_{1}\left(\frac{3 D}{P_{1} \cdot P_{1}}\right)$$
Meanwhile, $A$ is the projection of $P_{2}$ on $P_{1} H$, so
$$A = P_{1}\left(\frac{P_{1} \cdot P_{2}}{P_{1} \cdot P_{1}}\right) = P_{1}\left(\frac{D}{P_{1} \cdot P_{1}}\right)$$
Thus $H B=3 H A$, as desired.