Let $B$ be a $*$-algebra and $A\subseteq B$ a $*$-subalgebra. Let $p\in B$ be a projection such that
$$pBp=A.$$
Suppose we have a $*$-homomorphism $\phi:A\rightarrow\mathcal{B}(H)$, where $H$ is some Hilbert space.
Question: Can $\phi$ be extended to a $*$-homomorphism $\phi':B\rightarrow\mathcal{B}(H)$?
Not necessarily. Let $B= M_{2\times 2}(\Bbb C)$ and let $p=\begin{pmatrix}1&0\\0&0\end{pmatrix}$. Then $A$ are those matrices where only the $11$ component is non-zero. Let $\varphi: A\to \mathcal B(\Bbb C)=\Bbb C$ be the map sending such a matrix to its only non-zero component. This is a $*$-algebra morphism.
There can be no extension of this, because no $M_{2\times 2}(\Bbb C)$ admits no characters (meaning no $*$-algebra morphisms into $\Bbb C$). If you denote with $\varphi_{ij}$ the image of the matrix with a $1$ on the $ij$ component and else $0$ of such an extension, you have $(\varphi_{12})^2=0=(\varphi_{21})^2$ since this is an algebra morphism, implying $\varphi_{12}=\varphi_{21}=0$. But $\begin{pmatrix}0& 1\\ 1&0\end{pmatrix}$ is invertible, so either $\varphi_{12}+\varphi_{21}$ is invertible (not possible) or $\varphi$ is the zero morphism.