Martin's Axiom reduces to small ccc orders

Adjoin a largest condition if necessary. Starting with it, recursively form increasing sets $Q_n\subseteq P$ of size at most $\kappa$. To obtain $Q_{n+1}$, include all of $Q_n$; for every $q\in Q_n$ and every original dense $D$ choose one $d(q,D)\le q$ in $D$; and for every pair $q,r\in Q_n$ compatible in $P$, choose one common extension $s(q,r)\le q,r$. Put all these witnesses together with $Q_n$ into $Q_{n+1}$. AC supplies the simultaneous choices, and cardinal absorption preserves the size bound. Let $Q={\bigcup }_n{Q}_n$.

Every $D\cap Q$ is dense in $Q$ by the first closure requirement. Compatibility between members of $Q$ is reflected in $Q$ by the second: once both occur at some $Q_n$, their chosen common extension lies in $Q_{n+1}$. Hence every antichain of $Q$ is an antichain of the ccc order $P$, so $Q$ is ccc. A filter on $Q$ meeting the intersections generates an upward-closed directed filter in $P$ meeting every original $D$. Therefore the small-order restriction implies full $\mathrm{M}\mathrm{A}(\kappa )$; the converse is immediate.

Since $\mathrm{\mid }Q\mathrm{\mid }\le \kappa$, choose an injection $Q\hookrightarrow \kappa$ and transport the order to its image. The unused points of $\kappa$ are not forcing conditions; no duplicate largest elements are needed. This gives the fixed-domain coding used in bookkeeping without changing filters or ccc.