It's a little unclear what the three means are. I understand that the point of the question is get some help in making it clear, so let me go through some possibilities of what you meant and how you might state things more clearly.
$$\begin{align*}
\text{If }p &= \text{Mean}(x_1,2,3)+\text{Mean}(y_1,2,3)+\text{Mean}(z_1,2,3)\\
\\
&=\frac{x_1+2+3}{3}+\frac{y_1+2+3}{3}+\frac{z_1+2+3}{3}
\end{align*}$$
Then, this can be written more simply as
$$\begin{align*}
p &= \frac{x_1+y_1+z_1}{3}+\frac{2+2+2}{3}+\frac{3+3+3}{3}\\
\\
&= \frac{x_1+y_1+z_1}{3}+2+3\\
\\
&= \text{Mean}(x_1,y_1,z_1)+5
\end{align*}$$
On the other hand,
$$\begin{align*}
\text{if }p &= \text{Mean}(x_1,x_2,x_3)+\text{Mean}(y_1,y_2,y_3)+\text{Mean}(z_1,z_2,z_3)\\
\\
&=\frac{x_1+x_2+x_3}{3}+\frac{y_1+y_2+y_3}{3}+\frac{z_1+z_2+z_3}{3}
\end{align*}$$
Then I don't think you could write this more simply without choosing different notation. One way to do this would be to let
$$a_1=x_1\quad a_2=x_2\quad a_3=x_3$$
$$a_4=y_1\quad a_5=y_2\quad a_6=y_3$$
$$a_7=z_1\quad a_8=z_8\quad a_9=z_3$$
Then we have that
$$\begin{align*}
p &=\frac{x_1+x_2+x_3}{3}+\frac{y_1+y_2+y_3}{3}+\frac{z_1+z_2+z_3}{3}\\
\\
&= \frac{a_1+a_2+a_3}{3}+\frac{a_4+a_5+a_6}{3}+\frac{a_7+a_8+a_9}{3}\\
\\
&= \frac{1}{3}\cdot(a_1+a_2+\ldots+a_9)\\
\\
&= \frac{1}{3}\sum_{i=1}^9a_i
\end{align*}$$
You could also represent the sum of the three means more compactly using a double summation. We have that
$$\begin{align*}
p &= \text{Mean}(x_1,x_2,x_3)+\text{Mean}(y_1,y_2,y_3)+\text{Mean}(z_1,z_2,z_3)\\
\\
&=\frac{x_1+x_2+x_3}{3}+\frac{y_1+y_2+y_3}{3}+\frac{z_1+z_2+z_3}{3}
\end{align*}$$
Let's introduce the following notation:
$$a_{11}=x_1\quad a_{12}=x_2\quad a_{13}=x_3$$
$$a_{21}=y_1\quad a_{22}=y_2\quad a_{23}=y_3$$
$$a_{31}=z_1\quad a_{32}=z_8\quad a_{33}=z_3$$
Then we have that
$$\begin{align*}
p &= \frac{x_1+x_2+x_3}{3}+\frac{y_1+y_2+y_3}{3}+\frac{z_1+z_2+z_3}{3} \\
\\
&= \frac{a_{11}+a_{12}+a_{13}}{3}+\frac{a_{21}+a_{22}+a_{23}}{3}+\frac{a_{31}+a_{32}+a_{33}}{3} \\
\\
&=\frac{1}{3}\sum_{j=1}^3a_{1j}+\frac{1}{3}\sum_{j=1}^3a_{2j}+\frac{1}{3}\sum_{j=1}^3a_{3j}\\
\\
&= \frac{1}{3}\sum_{i=1}^3\sum_{j=1}^3a_{ij}
\end{align*}$$
I hope this helped. If you had something else in mind when you defined $p$, or if you had any questions about what I've wrote, then let me know.