Convert the given NFA to DFA I am trying to find an DFA for the regular language given by the expression $L\left( aa^{\ast }\left( a+b\right) \right)$.
First simplifying $L\left( aa^{\ast }\left( a+b\right) \right)$ we get
$L\left( aa^{\ast }\left( a+b\right) \right)$ $= L\left( a\right) L\left( a^{\ast }\right) L\left( a+b\right) $
Then I constructed an NFA for it , which is given below :

But I am not able to simplify the above NFA to a DFA as the state $q_1$ has two $\lambda$ transitions and I am not understanding how to deal with them .
See this page.
Essentially, create a new DFA $D$ in which the set of states $Q$ is the powerset of the set of states of your NFA $N$.
For $q \in Q$ and $a \in \Sigma$, add transition $\delta(q,a) = q'$ to $D$, if and only if, the set of states of $N$ that are reachable (in $N$) from at least one state in $q$ by reading character $a$ is exactly $q'$.
Mark a state of $q \in Q$ as a final state in $D$ if and only if $q$ contains a final state of $N$.
If $q_0$ is the initial state of $N$, then the initial state of $D$ is the set of states if $N$ that are reachable by $q_0$ using only $\varepsilon$-transitions.