= Solution
Prune an $\aleph_1$-<Suslin tree> as in part (a), then use its nodes as <forcing> conditions, with extensions stronger. Two conditions are compatible exactly when comparable, so the absence of uncountable <tree antichains> is the <forcing> <countable chain condition for forcing>. For each $\alpha<\omega_1$, the <set> $D_\alpha$ of nodes of height at least $\alpha$ is dense, by <well-pruned set-theoretic tree>.
If $2^{\aleph_0}>\aleph_1$, full <Martin's axiom> includes $\mathrm{MA}_{\aleph_1}$. It would provide a <filter in an ordered set> meeting all these <dense subsets of a forcing order>. Directedness makes that <filter in an ordered set> a <chain in a partial order>, and meeting every $D_\alpha$ makes its heights unbounded, producing an uncountable branch. This contradicts the Suslin-tree property. \b[A Suslin <set-theoretic tree> together with failure of <Continuum hypothesis> therefore implies failure of <Martin's axiom>.] This is the <Suslin-tree obstruction to Martin's axiom>.
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