In an unbiased $n-p$ junction electrons diffuse from \[n-region\] to $p-region$ because:
$(a)$electrons travel across the junction due to potential difference
$(b)$electron concentration in $n-region$ is more as compared to that of $p-region$
$(c)$holes in $p-region$ attract them
$(d)$only electrons move from $n-p$region and not the vice versa

Hint: We will use the basic property of $p-n$ junction for the explanation, that is, when $p-type$materials are connected to the $n-type$ material, they form a junction diode which at the point of contact has charges accumulated in between them which then transfers from one side to another. But, right after the rearrangement of these charges, there are no mobile charge carriers present to move from one side to another. This forms a barrier which further prohibits charge movement. This barrier is called depletion layer and the voltage required to again restart the charge transfer is known as Barrier potential.

Complete answer:
As we know, in $p-type$ materials, holes are the majority charge carriers and electrons are the minority charge carriers. Similarly, in $n-type$ materials, electrons are the majority charge carriers and holes are the minority charge carriers. Therefore, when these two types of materials are joined together, there is a concentration gradient in the number of electrons between the $n-type$ and the $p-type$ material. This concentration gradient between the two materials is the main reason why electrons diffuse from \[n-region\] to $p-region$.

Hence, option $(b)$ is the correct option.

Note:
It should be noted that electrons move under the influence of potential difference, only when there is a biasing. That is why option $(a)$ is incorrect.
Under Reverse Biasing electrons may flow from $p-n$ , that is why option $(d)$ is incorrect. Also, the attraction due to holes in $p-region$ on electrons in $n-region$ is minimal, so option $(c)$ is also incorrect.