As the small block does not slip over the prism, 
We can consider the mass of small block into prism. 

Where, $N$ is normal reaction 
and here  $(m +\hspace{0.17em}M)g \sin \theta$ is the only force in the line of inclination acting downwards 

$$\begin{gathered} F_{net}=ma \\ \Rightarrow \left(m+M\right)g\sin \theta =\left(m+M\right)a \\ \Rightarrow a=g\sin \theta \end{gathered}$$
$\therefore g \sin \theta$ will be its acceleration 
$\therefore$ option 1 and 2 are wrong 

Now FBD of block with respect to prism

Where, 
$mg \sin \theta$ is the pseudo force 
$f$ is force of friction 
$N$ is normal reaction 
Now writing equations of forces 
$\Rightarrow N + mg \sin \theta \sin \theta = mg$
$\Rightarrow N = mg - mg {\sin }^2\theta ...\left(1\right)$
and $f = mg \sin \theta \cos \theta$ 
$f = \mu \mathrm{N}$ 
$\Rightarrow \mu (mg - mg {\sin }^2\theta )=mg \sin \theta \cos \theta$
$\Rightarrow \mu (1 - {\sin }^2\theta ) = \sin \theta \cos \theta$ $\Rightarrow \mu {\cos }^2\theta = \sin \theta \cos \theta$ 
$\Rightarrow \mu = \tan \theta$ 
$\Rightarrow {\mu }_{min} = \tan \theta$