Let $$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$$ and $$B = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}, a, b \in N$$. Then

Compute the two possible products.

Left-multiply first:
$$AB=\begin{pmatrix}1&2\\3&4\end{pmatrix}\begin{pmatrix}a&0\\0&b\end{pmatrix}=\begin{pmatrix}1\cdot a+2\cdot 0 & 1\cdot0+2\cdot b\\3\cdot a+4\cdot 0 & 3\cdot0+4\cdot b\end{pmatrix}=\begin{pmatrix}a&2b\\3a&4b\end{pmatrix}$$

Right-multiply next:
$$BA=\begin{pmatrix}a&0\\0&b\end{pmatrix}\begin{pmatrix}1&2\\3&4\end{pmatrix}=\begin{pmatrix}a\cdot1+0\cdot3 & a\cdot2+0\cdot4\\0\cdot1+b\cdot3 & 0\cdot2+b\cdot4\end{pmatrix}=\begin{pmatrix}a&2a\\3b&4b\end{pmatrix}$$

For $$AB=BA$$ we equate the corresponding entries:

Upper-left: $$a=a$$ (always true).

Upper-right: $$2b=2a\;\Longrightarrow\;b=a$$.

Lower-left: $$3a=3b\;\Longrightarrow\;a=b$$.

Lower-right: $$4b=4b$$ (always true).

Thus the commutativity condition forces the single relation $$a=b$$. Any natural number $$a$$ produces a commuting matrix

$$B=aI_2=\begin{pmatrix}a&0\\0&a\end{pmatrix},\qquad a\in\mathbb N.$$

Since there are infinitely many natural numbers, there are infinitely many such diagonal matrices $$B$$.

Therefore, Option D is correct: there exist infinitely many $$B$$’s such that $$AB=BA$$.