 Solution

Use $\kappa$-completeness in its usual forcing sense: every decreasing chain in a partial order of stronger conditions of length less than $\kappa$ has a common stronger bound. Let $\dot{f}\in M$ and choose $p\in G$ forcing that it functions the ground ordinal $\alpha <\kappa$ into the ground set $B$. Below any stronger condition $r$, recursively decide each value of $\dot{f}$ in order. At a successor step the deciding conditions are dense, and at a limit stage use $\kappa$-completeness. After all $\alpha$ steps take another common bound. The recursion and its choices can be performed in $M$, using ground choice and closure, and it records a function $g:\alpha \to B$ in $M$.

Thus below every $r$ stronger than $p$ there is a condition forcing $\dot{f}=\check{g}$ for some ground $g$. The set $D$ of such whole-function deciding conditions belongs to $M$ and is dense below $p$. Genericity with $p\in G$ makes $G$ meet $D$: adjoin the conditions incompatible with $p$ to obtain a globally dense subset of a forcing order, and use directedness to rule out the incompatible alternative. A condition in $G\cap D$ then gives $f=g\in M$.

The reverse inclusion follows because ground functions remain functions with the same domain and values. ThereforeThis closed forcing adds no short ground-valued sequences argument needs density of complete decisions. A single arbitrarily constructed lower bound need not belong to $G$, and would not by itself prove the claim.