Given electric field

$\vec E = E_0(\hat x+\hat y)\sin(kz-\omega t)$

Since the phase is $kz-\omega t$, the wave propagates in the $+z$ direction.

Relation between $\vec E$, $\vec B$, and direction of propagation

For a plane electromagnetic wave in vacuum,

$\vec B = \frac{1}{c}\,\hat n \times \vec E$

where $\hat n$ is the unit vector in the direction of propagation.

Here,

$\hat n = \hat z$

so

$\vec B = \frac{1}{c}\hat z \times \vec E$

Compute the cross product

Substitute $\vec E$:

$\vec B = \frac{E_0}{c}\left[\hat z \times (\hat x+\hat y)\right]\sin(kz-\omega t)$

Using

$\hat z\times \hat x = \hat y, \qquad \hat z\times \hat y = -\hat x$

therefore,

$\hat z\times (\hat x+\hat y)=\hat y-\hat x=-\hat x+\hat y$

Hence,

$\vec B = \frac{E_0}{c}(-\hat x+\hat y)\sin(kz-\omega t)$

Match with options

This corresponds to Option D.

Check with stored answer

Stored correct answer: D

Derived answer: D

They match.