Srednicki writes: We can make this a little fancier by defining the unitary spacetime translation operator
$$ T(a) \equiv \exp(-iP^\mu a_\mu/ \hbar) $$
Then we have $$ T(a)^{-1} \phi(x) T(a) = \phi(x-a)$$
How do we get the second equation from the first equation?
It is unnecessary to use Hilbert space states to formally derive this result; it quickly follows from a useful result about the matrix exponential (which comes in very handy when one studies Lie algebras which, incidentally, is essentially what we're looking at here).
Let $X$ be any $n$-by-$n$ complex matrix, then we define the linear operator $\mathrm{ad}_X$ on the vector space of such matrices by $$ \mathrm{ad}_X Y = [X,Y] $$ for all $n$-by-$n$ complex matrices $Y$. Here $[\cdot, \cdot]$ denotes the commutator often called the adjoint operator. Then we have the following result: $$ e^XYe^{-X} = e^{\mathrm{ad}_X}Y $$ Now, if we formally apply this result to linear operators on the Hilbert space of a quantum field theory, then we obtain $$ T(a)^{-1}\phi(x)T(a) = e^{ia_\mu P^\mu/\hbar}\phi(x)e^{-ia_\mu P^\mu/\hbar} = e^{\mathrm{ad}_{ia_\mu P^\mu/\hbar}}\phi(x) = \sum_{k=0}^\infty \frac{\mathrm{ad}^k_{ia_\mu P^\mu/\hbar}\phi(x)}{k!} $$ Now we use the fact that for any field $\Phi$, we have $$ \mathrm{ad}_{ia_\mu P^\mu/\hbar}\phi(x)=\frac{ia_\mu}{\hbar} [P^\mu, \phi(x)] = \frac{ia_\mu}{\hbar}(i\hbar \partial^\mu)\phi(x) = -a_\mu \partial^\mu\phi(x) $$ Applying this result $k$ times and inserting it into the series expansion for the exponential written above, we obtain $$ T(a)^{-1}\phi(x)T(a) = \sum_{k=0}^\infty\frac{(-1)^k}{k!}(a_\mu\partial^\mu)^k\phi(x) $$ Now, we simply note that the right hand side is the Taylor expansion of $\phi(x-a)$. Explictly $$ \phi(x-a) = \phi(x) -a_\mu\partial^\mu\phi +\frac{1}{2}(a_\mu\partial^\mu)^2\phi(x) + \cdots + \frac{(-1)^k}{k!}(a_\mu\partial^\mu)^k\phi(x)+\cdots $$ and this gives the desired result.
[EDIT] Supposing a state basis $|e_i\rangle $, we are going to use the following notation for a state: $|A(x)\rangle = \sum_i A_i(x)|e_i\rangle$
An operator $O$ appying on $A$ gives then: $O|A(x)\rangle = \sum_i OA_i(x)|e_i\rangle$
For instance, $\partial_{\mu}|A(x)\rangle = \sum\partial_{\mu}A_i(x)|e_i\rangle$
Now, instead of working with operators, I think it is simpler to work with states $\left|A(x)\right\rangle$ and $\left|B(x)\right\rangle$ such as:
$$\left|B(x)\right\rangle = \Phi(x) \left|A(x)\right\rangle \tag{1}$$
This is true, of course, for $x-a$, that is:
$$\left|B(x - a)\right\rangle = \Phi(x- a) \left|A(x-a)\right\rangle \tag{2}$$
We know that:
$$P_{\mu}\left|A(x)\right\rangle = i \hbar \partial_{\mu}\left|A(x)\right\rangle.$$
So, we get: \begin{align} T(a)^{-1}\left|A(x)\right\rangle &= e^{\large iP_\mu a^\mu/ \hbar}\left|A(x)\right\rangle\\ &=e^{\large -a^{\mu}\partial_{\mu}} \left|A(x)\right\rangle\\ &= \left|A(x - a)\right\rangle \end{align}
The last equality is simply the Taylor series of $\left|A(x - a)\right\rangle$ at $x$, that is:
$$\left|A(x - a)\right\rangle = \left|A(x)\right\rangle - a^{\mu}\partial_{\mu} \left|A(x)\right\rangle + \frac{1}{2!} (a^{\mu}\partial_{\mu})^2\left|A(x)\right\rangle +\frac{(-1)^n}{n!}(a^{\mu}\partial_{\mu})^n\left|A(x)\right\rangle + \dots.$$
Now, applying $T(a)^{-1}$ to equation $(1)$, we get:
$$T(a)^{-1}\left|B(x)\right\rangle =T(a)^{-1}\Phi (x)\left|A(x)\right\rangle.$$
That is:
$$T(a)^{-1}\left|B(x)\right\rangle =T(a)^{-1}\Phi (x)T(a)T(a)^{-1}\left|A(x)\right\rangle.$$
So, we get:
$$\left|B(x - a)\right\rangle =T(a)^{-1}\Phi (x)T(a)\left|A(x - a)\right\rangle.$$
Looking at equation $(2)$, we finally get:
$$T(a)^{-1}\Phi (x)T(a) = \Phi (x-a)$$
