So I have found the Characteristic function of the variable $X_ \lambda$ to be:
$$\psi_{X_\lambda}(t) = \psi_{b(\lambda)(Y_\lambda-\lambda)}(t)=\mathbf Ee^{itb(\lambda)(Y_\lambda-\lambda)}=\mathbf E^{itY_\lambda b(\lambda)}e^{-it\lambda b(\lambda)}=e^{-it\lambda b(\lambda)} \mathbf Ee^{it Y_\lambda b(\lambda)} = e^{-it\lambda b(\lambda)} \psi_{Y_\lambda}(tb(\lambda))$$
Now as $Y_\lambda$ is a Poisson random variable we get: $$e^{\lambda(e^{itb(\lambda)}-1)-it \lambda b(\lambda)}$$
And this is where I am stuck at I don't know how to continue to establish convergence of $X_\lambda$ any hints?
Expanding on @Did's comment, we have $$ \psi_{X_\lambda}(t) = e^{\lambda\left(e^{itb(\lambda)}-1-itb(\lambda)\right)} $$ For sufficiently large $\lambda$, this is approximately equal to $$ e^{\lambda\left(-\frac12 t^2b^2(\lambda)\right)}.$$ Choosing $b(\lambda) = \lambda^{-\frac12}\sigma$ (where $\sigma>0$), we have $$\lim_{\lambda\to\infty}\psi_{X_\lambda}(t) = e^{-\frac12\sigma^2t^2}, $$ which implies that $$X_\lambda\stackrel{d}{\longrightarrow}\mathcal N(0,\sigma^2).$$