= Solution
Use the equivalent tree formulation of the <Suslin hypothesis>: there is no normal, well-pruned <Suslin tree> of height $\omega_1$, with countable levels and no uncountable chains or antichains. This is equivalent to the linear-order formulation that every complete dense order without endpoints satisfying the <countable chain condition for a linear order> is separable.
If such a tree $T$ existed, use its nodes as <forcing> conditions, with a higher extension stronger. It is <CCC> because its antichains are countable. For every $\alpha<\omega_1$, the set $D_\alpha=\{t:\operatorname{ht}(t)\ge\alpha\}$ is dense because the tree is well-pruned. The assertion $\mathrm{MA}_{\aleph_1}$ is precisely that a <CCC> <forcing> and a family of at most $\aleph_1$ dense sets admit a filter meeting all of them.
Apply it to these dense sets. A directed filter in a tree is a chain: two compatible nodes are comparable, since both lie among the well-ordered predecessors of a common extension. Meeting every $D_\alpha$ makes this chain cofinal, contrary to the defining absence of uncountable chains in a Suslin tree. Consequently
$$
\boxed{\mathrm{MA}_{\aleph_1}\ \Longrightarrow\ \text{Suslin's Hypothesis}.}
$$
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