Consider the following example that is used to understand the definition of periodicity property.
Why does it says that:
starting in state $1$, it is possible for the process to enter state $1$ only at times $2, 4, . . . ,$ so state $1$ has period $2$. The reason is that the player can break even (be neither winning nor losing) only at times $2, 4, . . . ,$ which can be verified by calculating $p_{11}^{(n)}$ for all $n$ and noting that $p_{11}^{(n)} = 0$ for $n$ odd.
I do not understand.
First off isn't $p_{11}^{(n)} = 0$ for all $n$, not just $n$ odd?
By definition $p_{11}^{(n)}=p\{X_{t+n}=1|X_t=1\}$
From the matrix P, $p_{11}^{(n)}=p\{X_{t+n}=1|X_t=1\}=0$, hence the probability will always be zero, no?
"First off isn't $p_{11}^{(n)} = 0$ for all $n$, not just $n$ odd?"
No. $\ p_{11}^{(2)}= p\left(1-p\right)\ne 0\ $. I suspect you're confusing state $0$ with state $1$. It is true that $\ p_{00}^{(n)}= 1\ $ for all $\ n\ $, so that state is aperiodic.
It is not correct that $\ p\{X_{t+n}=1|X_t=1\}=0\ $. $$ p\{X_{t+n}=1|X_t=1\}=\begin{bmatrix}0&1&0&0 \end{bmatrix}\boldsymbol{\mathrm{P}}^n\begin{bmatrix}0\\1\\0\\0 \end{bmatrix}\ ,$$ and $$ \boldsymbol{\mathrm{P}}^2=\begin{bmatrix} 1&0&0&0\\ 1-p&p\left(1-p\right)&0&p^2\\ \left(1-p\right)^2&0&p\left(1-p\right)&p\\ 0&0&0&1\end{bmatrix}\ .$$ It is the entry in the second row and second column of $\ \boldsymbol{\mathrm{P}}^n\ $that is $\ p_{11}^{(n)}\ $.