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

$\begin{array}{1 1}(a)\;There\;cannot\;exist\;any\;B\;such\;that\;AB=BA\\(b)\;There\;exist\;more\;than\;one\;but\;finite\;number\;of\;B's\;such\;that\;AB=BA\\(c)\;There\;exists\;exactly\;one\;B\;such\;that\;AB=BA\\(d)\;There\;exist\;infinitely\;many\;B's\;such\; that\;AB=BA\end{array}$

$A=\begin{bmatrix}1&2\\3&4\end{bmatrix}$
$B=\begin{bmatrix}a&0\\0&b\end{bmatrix}$
$AB=\begin{bmatrix}a&2b\\3a&4b\end{bmatrix}$
$BA=\begin{bmatrix}a&0\\0&b\end{bmatrix}\begin{bmatrix}1&2\\3&4\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}a&2a\\3b&4b\end{bmatrix}$
Hence $AB=BA$ only when $a=b$
$\therefore$ There can be infinitely many B's for which $AB=BA$
Hence (d) is the correct answer.