Girsanov theorem in CMS convexity derivation

Question

I am going through the derivation of CMS convexity from the notes of Lesniewski

There is a transformation from $T_p$ forward measure to annuity measure $Q$ as

$$ P(0,T_p)E^{Q_{T_p}}\left[S(T_0,T)\right]=A(0,T_0,T_n)E^Q\left[S(T_0,T)\frac{P(T_0,T_p)}{A(t,T_0,T_n)}\right] $$

where $A(t,T_0,T)=\sum_{1\le j \le n} \alpha_i P(t,T_i) $ is price of annuity at t paying $\alpha_i$ at $T_1,...,T_n$

Why is there an additional $P(T_0,T_p)$ term (zero coupon price of a bond maturing at $T_p$ and starting at $T_0$O in above equation?

Edit 1:I guess his notation is not clear. Right hand side can be written as $E[S(T_0;T_0,T)D(0,T_p)]$ where $D(t,T_p)=E\left[e^{-\int_t^{T_p}r dt}\right]$ and $S(T_0;T_0,T)$ is the swap spread at $T_0$ for the period from $T_0$ to $T$. $E[S(T_0;T_0,T)D(0,T_p)]=E^{Q_{T_p}}[S(T_p;T_0,T)]P(0,T_p)$ assuming $T_p$ is after $T_0$. At the same time $E[S(T_0;T_0,T)D(0,T_p)]=A(0;T_0,T_n)E^Q \left[S(T_0;T_0,T)\frac{D(T_0,T_p)}{A(T_0;T_0,T_n)}\right]$ through measure change

Answer 1

Note that $$\frac{dQ_{T_p}}{dQ}|_{T_0} = \frac{P(T_0, T_p)}{P(0, T_p)}\frac{A(0, T_0, T_n)}{A(T_0, T_0, T_n)}$$. Then $$E^{Q_{T_p}}\big(S(T_0, T_n)\big) = E^Q\bigg(S(T_0, T_n) \frac{P(T_0, T_p)}{P(0, T_p)}\frac{A(0, T_0, T_n)}{A(T_0, T_0, T_n)}\bigg) \\ = \frac{A(0, T_0, T_n)}{P(0, T_p)} E^Q\bigg(S(T_0, T_n) \frac{P(T_0, T_p)}{A(T_0, T_0, T_n)}\bigg).$$ That is, $$P(0, T_p)E^{Q_{T_p}}\big(S(T_0, T_n)\big) = A(0, T_0, T_n) E^Q\bigg(S(T_0, T_n) \frac{P(T_0, T_p)}{A(T_0, T_0, T_n)}\bigg). $$