With the below assumption:

• array1 has objects each of which always have prop data which is an array
• array2 also has objects and these objects will always have prop data which is an array as well
• the desired objective is to generate an array with objects from array1 whose data have at least one common element with any of the data of array2 objects.

this solution may work:

const findNeedleInHaystack = (needle = array1, haystack = array2) => (
    needle.reduce((acc, itm) => (
      haystack
      .map(x => x.data)
      .flat()
      .some(x => itm.data.includes(x))
        ? [...acc, itm]
        : [...acc]
    ), [])
);

Explanation

• iterate over the needle array (ie, array1) using .reduce
• use .map on haystack array (ie, array2) to separate the data from each object
• use .flat() to transform the 2-dimensional resulting array into 1-dimension
• use .some to see if any element in the 1-dimensional array is also present in itm's data array
• if found, concatenate itm to the result (using ... spread operator on acc - the aggregator/accumulator)
• if not found, skip concatenating itm.

Code Snippet

const array1 = [
  {
      id: '1',
      name: 'ron',
      data: ['p1']
  },
  {
      id: '2',
      name: 'lon',
      data: ['p2']
  },
{
      id: '3',
      name: 'voon',
      data: ['p4']
  }
];

const array2 = [
  {
      id: '1',
      name: 'fgr',
      data:['p1','p2','p3']
  },
  {
      id: '2',
      name: 'gone',
      data:['p1','p2','p3']
  }
];

const findNeedleInHaystack = (needle = array1, haystack = array2) => (
    needle.reduce((acc, itm) => (
    haystack.map(x => x.data).flat().some(x => itm.data.includes(x))
        ? [...acc, itm]
      : [...acc]
  ), [])
);

console.log(findNeedleInHaystack());