Answer
$\mathrm{C} < \mathrm{A} < \mathrm{B}$
Work Step by Step
The linear momentum of an object of mass $m$ moving with velocity $\vec{\mathrm{v}}$ is $\vec{\mathrm{p}}=m\vec{\mathrm{v}} \qquad$ 9-1
The kinetic energy of an object is $\displaystyle \mathrm{K}=\frac{1}{2}\mathrm{m}\mathrm{v}^{2}$
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$\displaystyle \mathrm{K}=\frac{\mathrm{m}\mathrm{v}^{2}}{2}\cdot\frac{\mathrm{m}}{\mathrm{m}}=\frac{\mathrm{m}^{2}\mathrm{v}^{2}}{2\mathrm{m}}=\frac{\mathrm{p}^{2}}{2\mathrm{m}}$
is a way of expressing kinetic energy with momentum and mass.
From here we express momentum in terms of K and m:
$\mathrm{p}^{2}=2\mathrm{m}\mathrm{K}$
$\mathrm{p}=\sqrt{2\mathrm{m}\mathrm{K}}$
$\mathrm{p}_{\mathrm{A}}=\sqrt{2\mathrm{m}\mathrm{K}}$
$\mathrm{p}_{\mathrm{B}}=\sqrt{2(4\mathrm{m})\mathrm{K}}=2\sqrt{2\mathrm{m}\mathrm{K}}=2\mathrm{p}_{\mathrm{A}}$
$\displaystyle \mathrm{p}_{\mathrm{C}}=\sqrt{2(\mathrm{m}/4)\mathrm{K}}=\frac{1}{2}\sqrt{2\mathrm{m}\mathrm{K}}=\frac{1}{2}\mathrm{p}_{\mathrm{A}}$
$\mathrm{C} < \mathrm{A} < \mathrm{B}$
