In the most general form, for arbitrary ideals over rings, this is false. In the ring $\mathbb{Z}$ let $I_k$ be generated by $2^k$ and let $J$ be generated by $3$. Then $I_k+J=\mathbb{Z}$ for all $k$ and so $\cap_{k=1}^\infty(I_k+J)=\mathbb{Z}$. But $\bigcap_{k=1}^\infty I_k=\lbrace0\rbrace$ and so $J+\bigcap_{k=1}^\infty I_k=J\ne\mathbb{Z}$.

In the most general form, for arbitrary ideals over rings, this is false. In the ring $\mathbb{Z}$ let $I_k$ be generated by $2^k$ and let $J$ be generated by $3$. Then $I_k+J=\mathbb{Z}$ for all $k$ and so $\cap_{k=1}^\infty(I_k+J)=\mathbb{Z}$. But $\bigcap_{k=1}^\infty I_k=\lbrace0\rbrace$ and so $J+\bigcap_{k=1}^\infty I_k=J\ne\mathbb{Z}$.

For an example in a $C^*$-algebra let $R=C[0,1]$ the continuous functions on $[0,1]$. Let $(a_k)$ be an enumeration of the rationals in $[0,1]$, and $I_k$ be the ideal of functions vanishing at $a_k$. Let $J$ be the ideal of functions vanishing at $1/\sqrt2$ say. Then $\bigcap_{k=1}^\infty I_k=\lbrace0\rbrace$ and $I_k+J=R$ for all $k$, and the rest proceeds as above.