The following statement is from Blackdar's book: If $J$ is a closed ideal of a $C^*$ algebra $A$,$\rho$ is a non-degenerate representation of $J$ on an $H$,and $(h_{\lambda})$ is an approximate unit for $J$,then for any $a$ in $A$,the net $(\rho(ah_{\lambda}))$ converges strongly in $L(H)$ to an operator $\rho_A(a)$,defing a representation of $A$ extending $\rho$ and we have $\rho_A(A)^{"}=\rho(J)^{"}$.
My question is: how to define the representation of $A$ such that it extends $\rho$ and why is $\rho_A(A)^{"}=\rho(J)^{"}$?
This is standard. Because $\rho$ is non-degenerate, you have that $\rho(J)H$ is dense. Define $\rho_A$ by $$ \rho_A(a)\rho(j)h=\rho(aj)h. $$ This is obviously a $*$-homomorphism. What is not immediately obvious is that it is well-defined. But if $\rho(j)h=0$, then \begin{align} \|\rho(aj)h\|^2&=\langle \rho(aj)^*\rho(aj)h,h\rangle=\langle\rho(j^*a^*aj)h,h\rangle\\[0.2cm] &\leq \|a\|^2\,\langle\rho(j^*j)h,h\rangle=\|a\|^2\|\rho(j)h\|^2=0. \end{align} The above estimate also shows that $\|\rho_A(a)\|\leq\|a\|$, so $\rho_A(a)$ is bounded and thus extends to all of $H$.