If you were to apply QPE to this unitary, what you will get, assuming you start with a proper eigenvector $|\Lambda\rangle$, is an estimation of $x$, if the associated eigenvalue $\Lambda$ is written as $\mathrm{e}^{2\mathrm{i}\pi x}$. Thus, to know what you will measure, you have to know what the eigenvalues of this unitary are.
There may be a faster way of doing this, but what you can do is to find the root of the characteristic polynomial $\chi$ associated to this matrix:
$$\chi = \begin{vmatrix}X-\cos\left(\frac\theta2\right)&\mathrm{e}^{\mathrm{i}\lambda}\sin\left(\frac\theta2\right)\\-\mathrm{e}^{\mathrm{i}\varphi}\sin\left(\frac\theta2\right)&X-\mathrm{e}^{\mathrm{i}(\varphi+\lambda)}\cos\left(\frac\theta2\right)\end{vmatrix}=X^2-2\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)\mathrm{e}^{\mathrm{i}\frac{\varphi+\lambda}{2}}+\mathrm{e}^{\mathrm{i}(\varphi+\lambda)}.$$
The roots of this polynomial are:
$$\Lambda=\mathrm{e}^{\mathrm{i}\frac{\varphi+\lambda}{2}}\left(\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)\pm\mathrm{i}\sqrt{1-\cos^2\left(\frac\theta2\right)\cos^2\left(\frac{\varphi+\lambda}{2}\right)}\right)$$
We can easily see that these roots lie, as expected, on the unit circle. We are thus interested by their argument. There are three cases to be dealt with:
- If $\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)>0$, then the eigenvalues are
$$\exp\left(\mathrm{2i\pi\frac{\frac{\varphi+\lambda}{2}+\arctan\left(\frac{\sqrt{1-\cos^2\left(\frac\theta2\right)\cos^2\left(\frac{\varphi+\lambda}{2}\right)}}{\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)}\right)}{2\pi}}\right)$$
and
$$\exp\left(\mathrm{2i\pi\frac{2\pi+\frac{\varphi+\lambda}{2}-\arctan\left(\frac{\sqrt{1-\cos^2\left(\frac\theta2\right)\cos^2\left(\frac{\varphi+\lambda}{2}\right)}}{\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)}\right)}{2\pi}}\right).$$
- If $\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)=0$, then the eigenvalues are $\exp\left(2\mathrm{i}\pi\frac{\varphi+\lambda+\pi}{4\pi}\right)$ and $\exp\left(2\mathrm{i}\pi\frac{\varphi+\lambda-\pi}{4\pi}\right)$.
- If $\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)<0$, then the eigenvalues are
$$\exp\left(\mathrm{2i\pi\frac{\frac{\varphi+\lambda+\pi}{2}+\arctan\left(-\frac{\sqrt{1-\cos^2\left(\frac\theta2\right)\cos^2\left(\frac{\varphi+\lambda}{2}\right)}}{\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)}\right)}{2\pi}}\right)$$
and
$$\exp\left(\mathrm{2i\pi\frac{\pi+\frac{\varphi+\lambda}{2}-\arctan\left(-\frac{\sqrt{1-\cos^2\left(\frac\theta2\right)\cos^2\left(\frac{\varphi+\lambda}{2}\right)}}{\cos\left(\frac\theta2\right)\cos\left(\frac{\varphi+\lambda}{2}\right)}\right)}{2\pi}}\right).$$
Thus, the value you get isn't only dependent on $\theta$, but also on $\varphi+\lambda$.