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I don't have enough reputation to comment but I wanted to add the following :

Like it was said in the comments in general this isn't possible but if you only care about the operator norm then this is always possible. In the sense that if $\pi$ is a unitary representation, there there exists a set of irreducible unitary representations $S$, such that if $a\in C^0_c(\Gamma)$ (or generally the $C^*$-algebra), then $||\pi(a)||=\sup_{\pi' \in S}||\pi'(a)||$.

Edit: Here is the proof :

1- You see the irreducible unitary representation $\pi$ as a representation $\pi:C^*\Gamma\to B(H)$.

2-The kernel $\ker(\pi)\subseteq C^*\Gamma$ is a closed $*$-ideal, and $A:=C^*\Gamma/\ker(\pi)$ is a $C^*$-algebra.

3-Since any injective $*$-morphism between $C^*$-algebras is an isometry, $||a||_A=||\pi(a)||_{B(H)}$ for any $a\in A$.

4-General theory of $C^*$-algebras tells that you that if $a\in A$, then $||a||_A$ is equal to $\sup ||\pi'(a)||$ where the sup is over all irreducible unitary representations of $A$ (Dixmier chapter 2)

5-By Dixmier (Chapter 3), set of irreducible untiary representations of $A$ is just a closed set of irreducible untiary representations of $\Gamma$. It is exactly those which factor through $\ker(\pi)$.

I don't have enough reputation to comment but I wanted to add the following :

Like it was said in the comments in general this isn't possible but if you only care about the operator norm then this is always possible. In the sense that if $\pi$ is a unitary representation, there there exists a set of irreducible unitary representations $S$, such that if $a\in C^0_c(\Gamma)$ (or generally the $C^*$-algebra), then $\|\pi(a)\|=\sup_{\pi' \in S}\|\pi'(a)\|$.

Edit: Here is the proof :

1- You see the irreducible unitary representation $\pi$ as a representation $\pi:C^*\Gamma\to B(H)$.

2-The kernel $\ker(\pi)\subseteq C^*\Gamma$ is a closed $*$-ideal, and $A:=C^*\Gamma/\ker(\pi)$ is a $C^*$-algebra.

3-Since any injective $*$-morphism between $C^*$-algebras is an isometry, $\|a\|_A=\|\pi(a)\|_{B(H)}$ for any $a\in A$.

4-General theory of $C^*$-algebras tells that you that if $a\in A$, then $\|a\|_A$ is equal to $\sup \|\pi'(a)\|$ where the sup is over all irreducible unitary representations of $A$ (Dixmier Chapter 2)

5-By Dixmier (Chapter 3), set of irreducible unitary representations of $A$ is just a closed set of irreducible unitary representations of $\Gamma$. It is exactly those which factor through $\ker(\pi)$.

I don't have enough reputation to comment but I wanted to add the following :

Like it was said in the comments in general this isn't possible but if you only care about the operator norm then this is always possible. In the sense that if $\pi$ is a unitary representation, there there exists a set of irreducible unitary representations $S$, such that if $a\in C^0_c(\Gamma)$ (or generally the $C^*$-algebra), then $||\pi(a)||=\sup_{\pi' \in S}||\pi'(a)||$.

Edit: Here is the proof :

1- You see the irreducible unitary representation $\pi$ as a representation $\pi:C^*\Gamma\to B(H)$.

2-The kernel $\ker(\pi)\subseteq C^*\Gamma$ is a closed $*$-ideal, and $A:=C^*\Gamma/\ker(\pi)$ is a $C^*$-algebra.

3-Since any injective $*$-morphism between $C^*$-algebras is an isometry, $||a||_A=||\pi(a)||_{B(H)}$ for any $a\in A$.

4-General theory of $C^*$-algebras tells that you that if $a\in A$, then $||a||_A$ is equal to $\sup ||\pi'(a)||$ where the sup is over all irreducible unitary representations of $A$ (Dixmier chapter 2)

5-By Dixmier (Chapter 3), irreducible untiary representations of $A$ is just a closed set of irreducible untiary representations of $\Gamma$. It is exactly those which factor through $\ker(\pi)$.

I don't have enough reputation to comment but I wanted to add the following :

Like it was said in the comments in general this isn't possible but if you only care about the operator norm then this is always possible. In the sense that if $\pi$ is a unitary representation, there there exists a set of irreducible unitary representations $S$, such that if $a\in C^0_c(\Gamma)$ (or generally the $C^*$-algebra), then $||\pi(a)||=\sup_{\pi' \in S}||\pi'(a)||$.

Edit: Here is the proof :

1- You see the irreducible unitary representation $\pi$ as a representation $\pi:C^*\Gamma\to B(H)$.

2-The kernel $\ker(\pi)\subseteq C^*\Gamma$ is a closed $*$-ideal, and $A:=C^*\Gamma/\ker(\pi)$ is a $C^*$-algebra.

3-Since any injective $*$-morphism between $C^*$-algebras is an isometry, $||a||_A=||\pi(a)||_{B(H)}$ for any $a\in A$.

4-General theory of $C^*$-algebras tells that you that if $a\in A$, then $||a||_A$ is equal to $\sup ||\pi'(a)||$ where the sup is over all irreducible unitary representations of $A$ (Dixmier chapter 2)

5-By Dixmier (Chapter 3), set of irreducible untiary representations of $A$ is just a closed set of irreducible untiary representations of $\Gamma$. It is exactly those which factor through $\ker(\pi)$.

I don't have enough reputation to comment but I wanted to add the following :

Like it was said in the comments in general this isn't possible but if you only care about the operator norm then this is always possible. In the sense that if $\pi$ is a unitary representation, there there exists a set of irreducible unitary representations $S$, such that if $a\in C^0_c(\Gamma)$ (or generally the $C^*$-algebra), then $||\pi(a)||=\sup_{\pi' \in S}||\pi'(a)||$.

I don't have enough reputation to comment but I wanted to add the following :

Like it was said in the comments in general this isn't possible but if you only care about the operator norm then this is always possible. In the sense that if $\pi$ is a unitary representation, there there exists a set of irreducible unitary representations $S$, such that if $a\in C^0_c(\Gamma)$ (or generally the $C^*$-algebra), then $||\pi(a)||=\sup_{\pi' \in S}||\pi'(a)||$.

Edit: Here is the proof :

1- You see the irreducible unitary representation $\pi$ as a representation $\pi:C^*\Gamma\to B(H)$.

2-The kernel $\ker(\pi)\subseteq C^*\Gamma$ is a closed $*$-ideal, and $A:=C^*\Gamma/\ker(\pi)$ is a $C^*$-algebra.

3-Since any injective $*$-morphism between $C^*$-algebras is an isometry, $||a||_A=||\pi(a)||_{B(H)}$ for any $a\in A$.

4-General theory of $C^*$-algebras tells that you that if $a\in A$, then $||a||_A$ is equal to $\sup ||\pi'(a)||$ where the sup is over all irreducible unitary representations of $A$ (Dixmier chapter 2)

5-By Dixmier (Chapter 3), irreducible untiary representations of $A$ is just a closed set of irreducible untiary representations of $\Gamma$. It is exactly those which factor through $\ker(\pi)$.