All right, so we know that the scalar field transforms as follows under the Lorentz transformation:
$$
\phi(x)\rightarrow\phi'(x)=\phi(\Lambda^{-1}x)
$$
as you have written. Note also that we have the following Lorentz transformation of the 4-vector:
$$
x^\mu\rightarrow x'^\mu=\Lambda^{\mu\;\;}_{\;\;\nu}x^\nu \ ,
$$
which may be rewritten as follows:
$$
x^\nu=(\Lambda^{-1})^{\nu\;\;}_{\;\;\mu}x'^\mu \ . \tag{1}
$$
Now, let's look closely at the transformation of the 4-derivative of the scalar field under the Lorentz transformation:
$$
\begin{align}
\partial_\mu\phi(x)\rightarrow\partial'_\mu\phi'(x)&=\frac{\partial}{\partial x'^\mu}\phi(\Lambda^{-1}x) \\
&=\frac{\partial x^\nu}{\partial x'^\mu}\frac{\partial}{\partial x^\nu}\phi(\Lambda^{-1}x) \\
&=\frac{\partial}{\partial x'^\mu}\left[(\Lambda^{-1})^{\nu\;\;}_{\;\;\rho}x'^\rho\right]\partial_\nu\phi(\Lambda^{-1}x) \\
&=(\Lambda^{-1})^{\nu\;\;}_{\;\;\rho}\delta^\rho_\mu\partial_\nu\phi(\Lambda^{-1}x) \\
&=(\Lambda^{-1})^{\nu\;\;}_{\;\;\mu}\partial_\nu\phi(\Lambda^{-1}x) \ .
\end{align}
$$
In the second line, we have used the chain rule, and in the third line, we have used Eq. (1). This shows that whenever we perform the Lorentz transformation, we have to transform not only the fields but also all 4-vectors. And doing this, we have obtained the result from Peskin&Schroeder.
