QM perturbation degenerate case $$
\begin{aligned}
0 &=\left(E-H_{0}-\lambda V\right)|l\rangle \\
&=\left(E-E_{D}^{(0)}-\lambda V\right) P_{0}|l\rangle+\left(E-H_{0}-\lambda V\right) P_{1}|l\rangle
\end{aligned}
$$
We next separate (5.2.2) into two equations by projecting from the left on (5.2.2) with $P_{0}$ and $P_{1}$,
$$
\begin{aligned}
&\left.\left(E-E_{D}^{(0)}-\lambda P_{0} V\right) P_{0}\left|l \rangle-\lambda P_{0} V P_{1}\right| l\right\rangle=0 \\
&\left.-\lambda P_{1} V P_{0}\left|l \rangle+\left(E-H_{0}-\lambda P_{1} V\right) P_{1}\right| l\right\rangle=0
\end{aligned}
$$
We can solve (5.2.4) in the $P_{1}$ subspace because $P_{1}\left(E-H_{0}-\lambda P_{1} V P_{1}\right)$ is not singular in this subspace since $E$ is close to $E_{D}^{(0)}$ and the eigenvalues of $P_{1} H_{0} P_{1}$ are all different from $E_{D}^{(0)}$. Hence we can write
$$
P_{1}|l\rangle=P_{1} \frac{\lambda}{E-H_{0}-\lambda P_{1} V P_{1}} P_{1} V P_{0}|l\rangle
$$
or written out explicitly to order $\lambda$ when $|l\rangle$ is expanded as $|l\rangle=\left|l^{(0)}\right\rangle+$ $\lambda\left|l^{(1)}\right\rangle+\cdots$.
$$
P_{1}\left|l^{(1)}\right\rangle=\sum_{k \notin D} \frac{\left|k^{(0)}\right\rangle V_{k l}}{E_{D}^{(0)}-E_{k}^{(0)}}
$$
Modern quantum mechanics JJ Sakurai page 299
My question is: there is operator $P_1$ right side of V on denominator of the fifth equation.I use the fourth equation to derive the fifth equation but I don't have $P_1$ right side of V in my calculation. Where does it come from?
 A: On the LHS of your 4th equation, the last term is $-\lambda P_1 V P_1 |l>$. Since $P_1^2=P_1$, this is equivalent to $-\lambda P_1 V P_1^2 |l>$. Then the rest is just a rearrangement of your 4th equation. you move the 1st term to the RHS, and multiply by the inverse operator in the parenthesis (its inverse exists in the $P_1$ subspace, as described in the text) to both sides, with the extra $P_1$ we just inserted included in the parenthesis. Now you get the 5th equation except for the first $P_1$ on the RHS. Act $P_1$ on both sides. On the LHS you get $P_1^2=P_1$ so nothing changes, on the RHS you get the RHS of the 5th equation.
It is not necessary to insert the $P_1$, i.e. in the denominator of your 5th equation, the last term can just be $-\lambda P_1 V$ as what you have got, rather than $-\lambda P_1 V P_1$ that appears in the text. However, including the inserted $P_1$  makes it clearer that the operator $-\lambda P_1 V$ is acting on the $P_1$ subspace, as seen from the last term on the LHS of the 4th equation.
