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**Hint:**The first rule to form factorization is to take common parts out at the start of the problem to work on the uncommon parts together. The common parts can be both constant and variable. We make a slight change in the sign to find the commons. Using the method, we form the factorization and find the solution.

**Complete step-by-step solution**

The given function is of the form $y\left( y-z \right)+9\left( z-y \right)$. The basic step of finding factorization is to take the common parts first. It can be both variable and constant.

In the case of $y\left( y-z \right)+9\left( z-y \right)$, we don’t have an exact common part. But after a slight change in the sign we can find the common.

$\begin{align}

& y\left( y-z \right)+9\left( z-y \right) \\

& =y\left( y-z \right)-9\left( y-z \right) \\

\end{align}$

Now we can take \[\left( y-z \right)\] from both parts.

$\begin{align}

& y\left( y-z \right)-9\left( y-z \right) \\

& =\left( y-z \right)\left( y-9 \right) \\

\end{align}$

The rule of factorisation is to form the multiplication form. The equation $y\left( y-z \right)+9\left( z-y \right)$ has been changed into $\left( y-z \right)\left( y-9 \right)$.

**The correct option is B.**

**Note:**We could have first simplified the multiplication already present in the problem and then have taken the common. But it would have been unnecessary as at the time of taking commons out the form would have returned to its previous form. So, to make things simple we just started from the given part to make the factorization. We can cross-check the given options to find the errors in them.

In case of $\left( y-z \right)\left( y+9 \right)$, it gives in expansion $\left( y-z \right)\left( y+9 \right)=y\left( y-z \right)+9\left( y-z \right)$ which is not our main equation.

In case of $\left( z-y \right)\left( y+9 \right)$, it gives in expansion $\left( z-y \right)\left( y+9 \right)=y\left( z-y \right)+9\left( z-y \right)$ which is not our main equation.

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