after cleaning up some open typos, I infer that you'll have probabilities $p_k\in(0,1)$
$P :=\left(\begin{matrix}
1-p_1 & p_1 & 0 & 0 & 0 & 0 & \dots\\
1-p_2 & 0 & p_2& 0 & 0 & 0 & \dots\\
1-p_3 & 0 & 0 & p_3 & 0 & 0 & \dots\\
1-p_4 & 0 & 0 & 0 & p_4 & 0 & \dots\\
1-p_5 & 0 & 0 & 0 & 0 & p_5 & \dots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots\end{matrix}\right)$
which is known as the (renewal) Age Matrix and is discussed in some detail in Feller vol 1 (3rd ed) Markov Chains chapter.
By inspection there is a single communicating class.
To verify transience/ recurrence, you merely need to check for state 1. And you don't return WP1 to state one iff
$\prod_{k=1}^\infty p_k \gt 0$.
Sums are easier to evaluate than products, so use the below identity, with $\delta_k = 1-p_k$
$\prod_{k=1}^\infty (1-1/2^k)$ converge to zero?
If the chain is transient, then you are done. If the chain is recurrent, then you may evaluate positive recurrence by simply checking whether the mean time until revisiting state 1 is finite. This may be computed directly as
$E\big[T\big] = \sum_{k=0}^\infty P\big(T\gt k\big) = 1 + \sum_{k=1}^\infty \prod_{j=1}^k p_j$