$\mathbf{\underline{Answer: A}}$
$\underline{\mathbf{Explanation:}}$
Total number of permutations of $\color{darkorange}{\mathbf{a, b, c, d}} = 4!= 24 $
Number of permutations of $\color{blue}{\mathbf{ab, c, d} }= 3!= 6 $
Number of permutations of $\color{blue}{\mathbf{a, bc, d}} = 3!= 6 $
Number of permutations of $\color{blue}{\mathbf{a ,b ,cd} }= 3!= 6 $
Number of permutations of $\color{green}{\mathbf b} \;\text{and}\; \color{green}{\mathbf c} = 2!= 2 $
Number of permutations of $\color{green}{\mathbf c} \;\text{and} \;\color{green}{\mathbf d} = 2!= 2 $
Number of permutations of $ \color{green}{\mathbf d} \;\text{and} \;\color{green}{\mathbf b}= 2! =2 $
Number of permutations of $ \color{green}{\mathbf{b,c}} \;\text{and} \; \color{green}{\mathbf d} = 1! = 1 $
Hence, required permutations are:
$\mathbf{4! - 3! - 3! - 3! + 2! + 2! + 2! - 1! = \underline{11}}$
$\therefore $ Option $\mathbf A$ is the correct answer.