A problem $L$ is $text{NP}$-complete if $L$ is in $text{NP}$, and $L$ is $text{NP}$-hard (that is, $Kleq_p L$ for all $Kin text{NP}$ ). Consider the following claims.

**Claim 1:** if $L$ is $text{NP}$-complete and $Lin text{P}$, then $text{NP} subseteq text{P}$ (that is, all problems in $text{NP}$ can be solved in deterministic polynomial time).

**Claim 2:** If $ A leq_p B$, then $Bin text{P} to A in text{P}$.

What you asked about is claim 1. To begin with, its correctness follows from claim 2. Indeed, if $L$ is $text{NP}$-complete, then for every $Ain text{NP}$, it holds that $Aleq_p L$. Thus, if we assume that $Lin text{P}$, then we get by claim 2 that $Ain text{P}$, and therefore $text{NP} subseteq text{P}$. So what is missing for you is the proof of claim 2, and a sketch of its proof goes as follows. If $M_f$ is a TM that computes a polynomial time reduction from $A$ to $B$, and $M_B$ is a TM that decides $B$ in polynomial time, then a TM $M_A$ that decides $A$ in polynomial time operates as follows. On input $x$, $M_A$ computes $y = M_f(x)$, then $M_A$ runs $M_B$ on $y$, and answers the same. What is left to show is that $M_A$ decides $A$ in polynomial time, and I leave that to you (use the fact that |y| is polynomial, and a composition of polynomials is a polynomial).