If two operators say $D$ and $B$ commutes then why a non-degenerate eigenfunction of operator $D$ is also an eigenfunction of operator $B$? I have following derivation which is not understandable for me and I am unable to understand it.Consider a eigenfunction-value equation $$D{\Psi}=d{\Psi}      $$
Now $B$ operates on above equation and gives$$BD{\Psi}=d(B{\Psi})      $$
since $d$ is a number. Furthermore, $B$ commutes with $D$, hence, we  can write above equation as $$D(B{\Psi})=d(B{\Psi})$$ above equation clearly shows that $B{\Psi}$ is a eigenfunction of operator D with same eigenvalue $d$.
HERE IS MY DOUBT IN STATEMENT 'since ${\Psi}$ is not degenerate with $B{\Psi}$ the eigenfunction $B{\Psi}$ must be a multiple of ${\Psi}$' i.e $$B{\Psi}=b{\Psi}$$ why this is necessary?
Explanation of this in my mind is since ${\Psi}$  and $B{\Psi}$ are  non-degenerate it must not give same eigenvalue,if above condition not met it is not possible to satisfy degeneracy of two eigen functions  . Is it can be written as following? $$D(B{\Psi})=db{\Psi}$$
my other question is if ${\Psi}$ is non degenerative then $B{\Psi}$ must be non degenerate?
 A: Consider an operator $D$ with some eigenfunction $\Psi$ and eigenvalue $d$:
$$D\,\Psi = d\,\Psi \quad. $$
If there is an operator, $B$, which commutes with $D$, then it follows that (as you showed) $B\, \Psi$ is also an eigenfunction of $D$ with the same eigenvalue $d$.
If $d$ is non-degenerate, then there is by definition only one eigenfunction (up to a multiplicative constant) which fulfills $D\,\Psi = d\, \Psi$. Hence, we must have that $B\,\Psi = b \Psi$ for some constant $b$.
Edit: Regarding your comment. How do we conclude that necessarily $B\,\Psi \overset{!}{=}b\,\Psi$ for some constant $b$ holds? Let us take a step back and consider an operator $A$ with two possible eigenstates. It could be the case that
$$A \,\psi_1  = a\,\psi_1 $$ and $$A\,\psi_2 = a\,\psi_2 \quad .$$ Actually, it is easy to see that if we chose $\psi_2 \equiv c\,\psi_1$ for some constant $c$, then the second equation follows trivially from the first one. If this is not the case, i.e. if $\psi_2 \neq c\, \psi_1$ for any constant $c$, then we say that the eigenvalue $a$ of $A$ is degenerated.
Back to our example, where we have assumed that $d$ is not degenerated. According to our above considerations this means that we cannot have two functions $\psi_1$ and $\psi_2$ which would fulfill
$$ D \,\psi_1  = d\,\psi_1$$
and $$ D\,\psi_2 = d\,\psi_2  $$
while $\psi_2 \neq c\, \psi_1$. But we found two functions, namely $\Psi$ and $B\,\Psi$ which fulfilled these equations. Hence, since at the same time it cannot hold that $B\,\Psi \neq b\,\Psi$, we find
$$  B\,\Psi = b\,\Psi \quad .$$
Hope this helped.
