To show $a \in V_{a+1}$ I'll use that $V_\alpha$ are transitive.
Claim 0: $V_\alpha$ are transitive, that is, if $a \in V_\alpha$ then $a \subset V_\alpha$
Proof: By transfinite induction over $\alpha$.
Case $\alpha = \emptyset$: We trivially have that for all $a$ in $V_\emptyset = \emptyset$ that $a \subset V_\emptyset$.
Case $\alpha = \beta + 1$ is a successor ordinal: Assume that for all $a \in V_\beta$ we have $a \subset V_\beta$ and let $b \in V_{\beta + 1} = P(V_\beta)$. Then for each $x$ in $b$ we have that $x \in V_\beta$. By assumption we have $x \subset V_\beta$, so for each $y \in x$ we have $y \in V_\beta$. Hence for each $y$ in $x$, $\{y \} \in P(V_\beta)$ and hence $\bigcup_{y \in x} \{y\} = x \in P(V_\beta) = V_{\beta + 1}$. So for each $x$ in $b$ we have $x \in V_{\beta + 1}$ which is the same as saying $b \subset V_{\beta + 1}$.
Case $\lambda$ is a limit ordinal: Assume that for all $\beta < \lambda$ we have $x \in V_\beta$ implies that $x \subset V_{\lambda}$ and let $b \in V_{\lambda} = \bigcup_{\beta < \lambda} V_\beta$. Then there exists a $\beta < \lambda$ such that $b \in V_\beta$. Then by assumption $b \subset V_\beta \subset \bigcup_{\beta < \lambda} V_\beta = V_{\lambda}$.
Now to answer the question:
Claim 1: $a \in V_{a + 1}$
Proof: By transfinite induction over $a$.
Case $a = \emptyset$: $a = \emptyset \in P(V_\emptyset) = P(\emptyset) = \{\emptyset\}$.
Case $a = b + 1$ is a successor ordinal: Assume that $b \in V_{b+1}$. We want to show $b + 1 \in V_{b + 2}$ where $b + 1 = b \cup \{b \}$. Since $b \in V_{b + 1}$ we have $\{b\} \in P(V_{b + 1}) = V_{b + 2}$. Since $V_\alpha$ are transitive we have $b \subset V_{b+1}$ and so for all $x$ in $b$, $x$ is also in $V_{b+1}$. Hence $\{x\} \in P(V_{b+1}) = V_{b+2}$ and so $b = \bigcup_{x \in b} \{x\} \in P(V_{b+1}) = V_{b+2}$. Hence $b \cup \{b\} \in V_{b+2}$ or in other words $a \in V_{a + 1}$.
Case $a = \lambda$ is a limit ordinal: Assume that we have $b \in V_{b+1}$ for all $b < \lambda$. We want to show that $\lambda \in V_{\lambda+1} = P(V_\lambda)$. By transitivity of $V_\alpha$ we have that if $c < d$ then $V_c \subset V_d$ hence for $b+1 < \lambda$ we get $V_{b+1} \subset V_\lambda$. By assumption, $b \in V_{b+1}$ hence $b \in V_\lambda$ for all $b + 1 < \lambda$ and hence $b \in P(V_{\lambda})$. So we have $\cup \{b \mid b + 1 < \lambda \} =\cup \{b \mid b < \lambda \} = \lambda \in P(V_{\lambda})$.
Claim 2: If $\alpha \in \beta$ then $V_\alpha \in V_\beta$
Proof:
I'll write $\alpha \in \beta$ as $\alpha < \beta$. We know that for $\alpha < \beta$ we have $V_\alpha \subset V_\beta$.
If $\beta$ is a successor ordinal then there is a $\gamma$ such that $V_\beta = P(V_\gamma)$ for $\gamma + 1 = \beta$. If $\alpha = \gamma$ then $V_\alpha \in V_\beta$. If $\alpha < \gamma$ then $V_\alpha \in V_{\alpha + 1} \subseteq V_\gamma \subset V_\beta$.
If $\beta > 0$ is a limit ordinal then $V_\beta = \bigcup_{\gamma < \beta} V_\gamma$. Since $\beta$ is a limit ordinal we have $\alpha < \alpha + 1 < \beta$ and hence $V_\alpha \in P(V_\alpha) = V_{\alpha + 1} \subset V_\beta$.