Is the universal representation of a C*-algebra same as this one? Let $A$ be a $C^*$-algebra and denote the state space of $A$ by $S(A)$. Since every $\rho\in S(A)$ gives rise to a GNS representation $(\pi_{\rho},H_{\rho})$, we define the universal representation of $A$ as the pair $(\pi,H)$, where $\pi=\sum_{\rho\in S(A)}\oplus \pi_{\rho}$ acting upon $H=\oplus_{\rho\in S(A)}H_{\rho}$. This representation is faithful.
We can also define another faithful representation of $A$. Each nonzero $a\in A$ gives rise to a state $\varphi_a\in S(A)$ such that $\varphi_a(a^*a)=\|a\|^2$. Corresponding to each $\varphi_a\in S(A)$, we get a GNS representation $(\pi_a,H_a)$, and taking the direct sum of all these (for nonzero $a$'s) we get the representation $\Phi=\sum_{a\in A\setminus 0}\oplus \pi_a$ acting on $K=\oplus_{a\in A\setminus 0}H_a$.
So I am curious about few things - 


*

*Are these two representations unitarily equivalent?

*What if, instead of taking the direct sum of all representations corresponding to nonzero elements in $A$, we take the direct sum of all representation corresponding to norm one elements in $A$? 

*Does the universal representation has some universal property, for example, like factorization?

 A: There is a level of ambiguity in your question. Let $A=\mathbb C$, and consider the following two representations $\pi_1:A\to B(\mathbb C)=\mathbb C$ and $\pi_2:A\to M_2(\mathbb C)$, where $\pi_1(a)=a$ and $\pi_2(a)=aI_2$. These two representations are isomorphic (you won't be able to distinguish them by algebraic properties) but they are not unitarily equivalent (not even approximately) because the Hilbert spaces on which they represent have different dimensions. 
The universal representation is unique up to isomorphism, but not up to (approximate) unitary equivalence because of multiplicity issues like above. You are taking every state on $A$ to build the representation but you could have taken every positive functional (i.e. every positive multiple of every state) and you might be changing multiplicities because of that. This problem also appears when choosing your $\Phi$, where you take all $a$ or just those with $\|a\|=1$.
To further complicate the problem, there is another ambiguity in your $\Phi$, in that you have to choose a $\varphi_a$ for every $a$, and there is no canonical way of doing it. 
(Approximate) unitary equivalence of representations is given (due to work of Voiculescu and later Hadwin) by equality of the rank of the image of every single element in the algebra. Since multiplicity issues will change the rank, as in the example above, one would have to be very careful with multiplicities. 

Defining the universal representation the way you did is certainly very common, but it hides most of the reasons to construct it. 
Takesaki defines a representation $\pi:A\to H$ to be universal if for any representation $\rho:A\to H$ there exists a $\sigma$-weak continuous $*$-homomorphism $\tilde\rho:\pi(A)''\to\rho(A)''$ such that $\rho=\tilde\rho\circ\pi$. 
A universal representation is unique up to isomorphism: if $\pi_1,\pi_2$ are universal representations for $A$, then there exists a $\sigma$-weak continuous $*$-isomorphism $\gamma:\pi_1(A)''\to\pi_2(A)''$ such that $\gamma\circ\pi_1=\pi_2$. 
Then one constructs the representation direct sum of all GNS representations of all states (or positive functionals) and prove that it is universal, and for most purposes it becomes the universal representation. 
The proof that the direct sum of the GNS representations of all the states is universal depends crucially on having all states available. So I don't immediately see how to show that your $\Phi$ is universal. 
