= Solution
The high-pressure <lattice constant> follows from its first $(100)$ reflection:
$$
a_P=\frac{0.154\,\mathrm{nm}}{2\sin14.85^\circ}=0.30044\,\mathrm{nm}.
$$
Conserve the number of NaCl <formula units>. The low-pressure conventional <unit cell> contains four, whereas the high-pressure cell contains one. The <volume> per <formula unit> changes from $a_0^3/4$ to $a_P^3$, so
$$
\boxed{\frac{V_P}{V_0}=4\left(\frac{a_P}{a_0}\right)^3\simeq0.605.}
$$
The sample contracts by about $39.5\%$. Comparing $a_P^3$ directly with $a_0^3$ would compare different amounts of material.
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